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Unformatted text preview: Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures we can solve for y to obtain Solutions to The Grammar of Fixed Income Securities
1. Part Part Part Part Part Part a. b. c. d. e. f. $100/(1.20)3 = $57.87. $100/(2)3 = $12.50. $100/(1)3 = $100. $100/(1.10)6 = $56.45. $100/(1.05)12 = $55.68. $100/e.203 = $54.88. y= So yB = and yA = 2nLN nSN 1 + c/2 1 . P/100 1 + .09125/2 2 184  1 = 2.991185% 105 103.6776495/100 2 184 1 + .09125/2  1 = 2.778233% . 105 103.7401495/100 2. Remember we are trying to find r such that the growth in our initial investment is the same as the growth from an alternative investment providing a rate of r(m). That is, find r Part a. Part b. Part c. Part d. such that er = (1 + r(m)/m)m so r = m ln(1 + r(m)/m). m = 1: ln(1.04) = 3.922%. m = 1: ln(1.20) = 18.232%. m = 4: 4 ln(1.05) = 19.516%. m = 1: ln(2) = 69.315%. Part b. Ignoring for the moment any restrictions on lot size, we note that a cash flow of $100 to be received on 12/31/92 and purchased on 9/17/92 is available either in the form of $100 par amount of the 12/31/92 bill or in the form of $100/(1+.09125/2) par amount of the 9 1/8's of 12/31/92. (Indeed, since $100 par of the 9 1/8's will pay off $(100 + 9.125/2) on 12/31/92, $100/(1 + .09125/2) par amount of the 9 1/8's will pay off $100.) Can we buy this cash flow low through one instrument and sell it high through the other? We must see if the asked price of one of the instruments is lower than the bid price of the other. For the T bill, we have computed from Example 14 the asked price: A PBILL 99.17 . In addition, we have
B PBILL = 100 1  d nSM 360 = 100 1  .0288 105 360 99.15 . 3. Part a. Settlement is 9/17/92, the next coupon date is 12/31 the last coupon date is 6/30/92, therefore nSN = 105, nLS = 79, and nLN = 184. In addition, c = .09125. Accrued interest is 9.125 79 a= = 1.958899457 , 2 184 B so the full bid price is P , is 101 23/32 + 1.958899457 = 103.6776495 , and the full asked price is P A , is 101 25/32 + 1.958899457 = 103.7401495 . Since P = 100 1 + c/2 nSN , 1 + y/2 nLN
Page 1333 For the bond we must scale the full prices down to represent $100 payoff on 12/31/92, or in other words, multiply by 1/(1 + .09125/2). Using the full prices computed in part (a) we have 103.7401495 A 99.21 , scaled PBOND = (1 + .09125/2) 103.6776495 B 99.15 . scaled PBOND = (1 + .09125/2) If we buy the bond and short the bill, we must pay $99.21 for the bond and receive $99.15 from the bill sale, and so we lose 6/. If we buy the bill and short the bond, c we pay $99.17 for the bill and receive $99.15 from the bond sale, so we lose 2/. Thus, c there is no arbitrage opportunity. 4. The settlement date is 9/17/92. The price of 1026 or 10 26/32 = 10.8125 is both
Chapter 29: Solutions To Questions, The Grammar of Fixed Income Securities Page 1334 Chapter 29: Solutions To Questions, The Grammar of Fixed Income Securities Adventures in Debentures Adventures in Debentures the flat and full price of $100 par amount since a zerocoupon bond has zero accrued interest. Recall our formula for the price of a Treasury bond given yield: P = 100 (1 + y/2)nSN /nLN
TNM Settlement date: 10/13/94. Last coupon date: 10/2/94. Next coupon date: 4/2/95. t=0 c/2 (1 + y/2)t + 1 (1 + y/2)TNM . nLN = 180. nSN = 360(95  94) + 30(4  10) + (2  13) = 169. nLS = 360(94  94) + 30(10  10) + (13  2) = 11. When c = 0, the formula reduces to P = 100 (1 + y/2)TNM +(nSN /nLN ) which we can invert to solve for y in terms of P : y=2 100 P
1/(TNM +(nSN /nLN )) Full Price = 1 . 100 (1 + .07/2)169/180 .09/ 1 .09 2 + + = 102.114144. 2 (1 + .07/2) 1 + .07/2 Accrued Interest = 4.5 The next coupon date is 11/15/92 and the last coupon date is 5/15/92 so nSN = 59, nLN = 184, and TNM = 58. Thus, the yield you get on your STRIPs is y=2 100 10.8125
1/(58+59/184) 11 180 = 0.275000. Flat or Quoted Price = 102.114144  0.275000 = 101.839144. or in 32nd 's: 101 : 27.  1 = 7.7757522% . Part c. 5. Part a. Settlement date: 10/11/94. Last coupon date: 7/31/94. Next coupon date: 1/31/95. nSN = 112 nLN = 184 nLS = 72. Full Price = 100 1 + .05 = 102.504244. 1 + (.04 112/184) 72 184 = 1.956522. Settlement date: 10/11/94. Last coupon date: 7/28/94. Next coupon date: 1/28/95. nLN = 180. nSN = 360(95  94) + 30(1  10) + (28  11) = 107. nLS = 360(94  94) + 30(10  7) + (11  28) = 73. Accrued Interest = 5 Full Price = 100 (1 +
.085/ )107/180 2 .10/ .10 1 2 = 103.135832. + + 2 (1 + .085/2) 1 + .085/2 Flat or Quoted Price = 102.504244  1.956522 = 100.547723. or in 32nd 's: 100 : 18. Accrued Interest = 5 73 180 = 2.027778. Flat or Quoted Price = 103.135832  2.027778 = 101.108054. Part b.
Chapter 29: Solutions To Questions, The Grammar of Fixed Income Securities Page 1335 or in 32nd 's: 101 : 03. Chapter 29: Solutions To Questions, The Grammar of Fixed Income Securities Page 1336 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Part c. For time period 0.0, the value of this constant coupon bond is: Solutions to Bond Valuation Using Synthetics
1. Part a. For time period 0.0, the value of this constant coupon bond is:
20 20 9,158.04 = (1009.09
t=1 etrt ) + 9363.03e20.109444 For time period 0.5, the value of this constant coupon bond is:
20 10,000.03 = (1009.09
t=1 etrt ) + 9363.03e20.099444 9,658.51 = (1009.09
t=1 e(t.5)rt.5 ) + 9363.03e19.5.109297 For time period 0.5, the value of this constant coupon bond is:
20 For time period 1.0, the value of this constant coupon bond is: e(t.5)rt.5 ) + 9363.03e19.5.099297 19 10,496.43 = (1009.09
t=1 9,173.27 = (1009.09 For time period 1.0, the value of this constant coupon bond is:
19 t=1 etrt ) + 9363.03e19.109132 10,004.32 = (1009.09
t=1 etrt ) + 9363.03e19.099132 Part d. Part b. For time period 0.0, the value of this constant coupon bond is:
20 10,964.66 = (1009.09
t=1 etrt ) + 9363.03e20.089444 Passage of Time (in Years) 0.0 0.5 1.0 For time period 0.5, the value of this constant coupon bond is:
20 ShortTerm Rate: e(t.5)rt.5 ) + 9363.03e19.5.089297 11,454.33 = (1009.09
t=1 7.00% 8.00% 9.00% $10,964.66 $10,000.03 $9,158.04 $11,454.33 $10,496.43 $9,658.51 $10,952.61 $10,004.32 $9,173.27 For time period 1.0, the value of this constant coupon bond is:
19 10,952.61 = (1009.09
t=1 etrt ) + 9363.03e19.089132 Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1337 Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1338 Adventures in Debentures Adventures in Debentures Performance Surface: 20 Yr Coupon Bond $11,500.00 $11,000.00 Dollar Value $10,500.00 A and B. Assume that we buy 1,000 C bonds. This costs us 960,000 dollars. We will receive 100,000 dollars in year 1 and 1,100,000 dollars in year 2. Now consider selling 100 type A bonds and 1,100 type B bonds. This will give us 973,500 dollars. Subtracting the 960,000 dollars it cost us to purchase the 1,000 type C bonds, we see that we have a 13,500 arbitrage profit as long as we do not owe anything in the future. In year 1 we will owe 100,000 dollars to those who purchased our 100 type A bonds. The coupon payments we receive from our 1,000 type C bonds (which we bought) total 100,000 dollars. We can therefore pay off the short position on the 100 type A bonds. In year 2 we will owe the holders of the 1,100 type B bond the 1,100,000 dollars but this is exactly what we receive from the 1,000 type C bonds. Thus, in the future we owe nothing and we receive 13,500 dollars today. Obviously, there is no reason to do this on the small scale we have just described. If we purchase twice the number of C bonds and sell short twice the number of A and B bonds, we make an arbitrage profit of 27,000 dollars. Avarice and unbridled greed should push us even further in this direction. Now we turn to a more algebraic approach to the solution, and we begin by formulating an algebraic statement of the desired position. For the moment assume we want the cash inflow in year 0 (CF0) to be +13,500  any positive number would be acceptable. We want the cash flow in years 1 and 2 (that is, CF1 and CF2) to be zero. If we could find such a position, we have found a classic arbitrage. Thus, we want: CF 0 : 935NA  800NB  960NC = 13,500 0NB + 100NC = 0 CF 1 : 1000NA + 0 CF 2 : 0NA + 1000NB + 1100NC = After solving the above three equations in three unknowns, the solution is: NA = 100 NB = 1100 NC = 1000 . $10,000.00 1 $9,500.00 0.5 Passage of Time $9,000.00 0.07 0.08 0 0.09 Term Structure Indicator 2. There are two alternative, but equivalent, approaches to this problem. The first solution that follows is (perhaps) more intuitive, but the solution tends to ramble. The second method provides a clear (although algebraic) path toward generating a solution. It is immediately obvious that if bonds A an B are priced correctly, then 0 d1 = .935 and 0 d2 = .800. If this is the discount function, then bond C should be selling for .935 100 + .800 1,110 = 973.5 Bond C is only selling for 960 dollars. This inconsistency implies that there is an arbitrage opportunity. It would appear that bond C is under priced relative to bonds A and B. To take advantage of this we can purchase bond C and go short bonds
Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1339 In the equation labeled CF0, the right hand side was arbitrary. If we found a solution with any positive number on the right hand side of the CF0 equation, we found an arbitrage strategy that generated positive net cash flow today and no future net cash flows. 13,500 was a convenient choice1 given our first approach at the beginning of the solution; clearly, both formulations provide the same numerical answer. If we use a different value on the right hand side of CF0, then we would get a different combination of assets A, B, and C, but it would still be an arbitrage. 3.
1 Another convenient choice for the right hand side of equation CF0 is 1. After finding the numerical solution for an arbitrage profit of one dollar, it is simple to scale up the arbitrage strategy to generate any level of desired profit. Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1340 Adventures in Debentures Adventures in Debentures There are two alternative, but equivalent, approaches to this problem. The first solution that follows is (perhaps) more intuitive, but the solution tends to ramble. The second method provides a clear (although algebraic) path toward generating a solution. There is an arbitrage opportunity if these four bonds are selling at these prices. One way to discover the opportunity involves inferring the discount that is consistent with the prices of some of the bonds, as we did above. If this discount function is not consistent with the prices of all bonds, then there is an arbitrage opportunity. Observe first that bond B can be used to determine 0 d1 , the price paid today for a dollar to be delivered in one year. We have 93 = (0 d1 )100 or 0 d1 = .93. This is the value of 0 d1 that justifies the price of bond B. Now look at bond C. If 0 d1 were equal to .93, then 0 d2 , the price paid today for a dollar to be delivered in 2 years, would have to satisfy 92.85 = (.93)5 + (0 d2 )105 for bond C to be priced correctly. The 0 d2 that satisfies this is .84. Turn now to bond A. If 0 d1 = .93 and 0 d2 = .84, then A is correctly priced only if 0 d3 (the current value of a dollar in 3 years) satisfies 100.20 = (.93)10 + (.84)10 + (0 d3 )110 . For this to hold, 0 d3 must equal .75. Now ask what the price of bond D would be if 0 d1 = .93, 0 d2 = .84 and 0 d3 = .75. We have Justified Price = (.93)20 + (.84)20 + .75(120) = 18.6 + 16.8 + 90.0 = 125.4 . This implies that bond D is under priced relative to the other three bonds. This means that to exploit the arbitrage opportunity we will want to buy D and issue (short) bonds A, B, and C. Note first that for every D bond we buy we receive 120 dollars in year 3. An A bond, if shorted, requires that we pay out 110 dollars in year 3. This means we can issue or short 12 A bonds for every 11 D bonds we purchase and not owe anyone anything in year 3. If we short 12/11 A bonds for every D bond we purchase, we will receive in year 2 twenty dollars from the D bond we own but we will only owe 10.90909 dollars on our short position in A bonds. This leaves 9.09091 dollars. We can therefore short 9.09091/105 = .08658 C bonds and payoff what is owed on that short position in year 2 with the 9.09091 we receive. Now look at our position in year 1. We receive in one year 20 dollars for every D bond purchased but we owe 10.90909 dollars on the 12/11 A bonds we have shorted and .4329 on the .08658 C bonds we have shorted. This leaves 8.658 dollars. This means we can short .08658 B bonds. The position we have taken is given by the simple recipe: for every D bond purchased, short 1.090909 A bonds, .08658 B bonds, and .08658 C bonds. If this is done, nothing is owed or received in the future. However, we make (free
Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1341 lunch) money today. The D bond costs 121.2. From the 1.090909 A bonds we short we receive 109.3090 dollars. From the .08658 shorted B bonds we receive 8.051948 dollars and from the .08658 shorted C bonds we receive 8.0388961 dollars. The total of the short sale receipts is therefore 125.40 dollars. Since we only pay out 121.2 dollars for the D bond, we make $4.20 for each D bond that we buy.2 Of course, we need not purchase only one D bond; we can make much greater profits (if prices remain as they are) by purchasing, say, 10,000 D bonds. Indeed, if prices do not change, one can make infinite profits by taking an infinite position. Of course, this is an absurdity  prices must change. If the D bond's price rises to 125.4 the arbitrage opportunity disappears. Only then can the market be in equilibrium. Otherwise there is infinite demand for the under priced bond. Now we turn to a more algebraic approach to the solution, and we begin by formulating an algebraic statement of the desired position. For the moment assume we want the cash inflow in year 0 (CF0) to be +4.20  any positive number would be acceptable. We want the cash flow in years 1, 2, and 3 (that is, CF1, CF2, and CF3) to be zero. If we could find such a position, we have found a classic arbitrage. Thus, we want: CF 0 : 100.20NA CF 1 : 10.00NA CF 2 : 10.00NA CF 3 : 110.00NA  93.00NB + 100.00NB + 0.00NB + 0.00NB  92.85NC + 5.00NC + 105.00NC + 0.00NC  121.20ND + 20.00ND + 20.00ND + 120.00ND = = = = 4.20 0.00 0.00 0.00 After solving the above four equations in four unknowns, the solution is: NA = 1.09091 NB = 0.08658 NC = 0.08658 ND = 1 . In the equation labeled CF0, the right hand side was arbitrary. If we found a solution with any positive number on the right hand side of the CF0 equation, we found an arbitrage strategy that generated positive net cash flow today and no future net cash flows. 4.20 was a convenient choice3 given our first approach at the beginning of the solution; clearly, both formulations provide the same numerical answer. If we use a different value on the right hand side of CF0, then we would get a different combination of assets A, B, C, and D, but it would still be an arbitrage.
2 3 If you find the fractional units used in this explanation troublesome, consider purchasing 11,550 D bonds and shorting 12,600 A bonds, 1,000 B bonds and 1,000 C bonds. This yields a tidy arbitrage profit of 48,510 dollars. Check to see that nothing is owed on this position in the future. Another convenient choice for the right hand side of equation CF0 is 1. After finding the numerical solution for an arbitrage profit of one dollar, it is simple to scale up the arbitrage strategy to generate any level of desired profit. Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1342 Adventures in Debentures Adventures in Debentures 4. Part a. A dedicated portfolio consisting of the zeros would contain: Maturity Number of Zeros Price Total Cost 1 0 0.9091 0.0000 2 20 0.8000 16.0000 3 45 0.7143 32.1435 4 80 0.6250 50.0000 98.1435 So, assume you are going to replicate Bond C with Bonds A and B. Then, you will have the cash flows of NA Bond A plus cash flows of NB Bond B equal to the cash flows of Bond C. That is, the cash flows in year t (i.e., CF t) are: CF 1: CF 2: CF 3: 5NA 5NA 105NA + 4NB + 4NB + 104NB = 7 = 7 = 107 . Note that the second equation is redundant given the first. Subtract 21 times CF 1 from CF 3, Part b. A dedicated portfolio consisting of the coupon bonds would contain: Bond Number of Bonds Price Total Cost A  0.0500 98.0874  4.9044 B 10.2055 99.0805 1,011.1660 C 10.0000 104.4055 1,044.0550 D 0.6576 106.6940 70.1620 98.1466 105NA + 104NB 105NA  84NB 20NB NB Substitute into equation for CF 1: = 107 = 147 = 40 = 2 . 5NA + 4(2) = 7 5NA = 15 NA = 3 . So 3 Bond A and (2) Bond B gives the same cash flows as Bond C. The cost of 3 units of Bond A and shorting 2 units of Bond B is 3(100)  2(95) = 110 < 112. To earn an arbitrage profit of $2, short 1 unit of Bond C, buy 3 units of bond A, and short 2 units of Bond B. Now we turn to a more direct approach to the solution, and we begin by formulating an algebraic statement of the desired arbitrage position. We want the cash inflow in year 0 (CF0) to be +2.00. We want the cash flow in years 1, 2, and 3 (that is, CF1, CF2, and CF3) to be zero. If we could find such a position, we have found a classic arbitrage. Thus, we want: CF 0 : 100NA CF 1 : 5NA CF 2 : 5NA CF 3 : 105NA  95NB + 4NB + 4NB + 104NB  112NC + 7NC + 7NC + 107NC = 2.00 = 0 = 0 = 0 Notice that the cost of the two dedicated strategies in parts a and b are the same (within rounding error). However, the solution using zeros is easier to calculate and easier to implement in practice. There are some institutions that could not implement part b because it involves shorting Bonds A and B. 5. There are two alternative, but equivalent, approaches to this problem. The first (and perhaps more intuitive) solution tries to find a synthetic and then compares the cost of the synthetic to the actual security. If the costs of the actual and the synthetic are different, then arbitrage is possible by buying the cheap and selling the expensive. The second method is similar, but more direct. It looks for an arbitrage strategy without doing the intermediate step of constructing a synthetic. To find an arbitrage opportunity, you must replicate the cash flows of one of the bonds with a portfolio of the other two. Pick two bonds, and try to form a portfolio that replicates the payoffs of the other one. For this problem, it will not matter which two bonds you pick because you will still find an arbitrage opportunity.4
4 There may be situations where the selection of the securities to be placed in the synthetic matter. For example, if you have only one security mispriced relative to a small subset of the total securities, then you might have to examine more than one case. Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1343 Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1344 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Since the equations for CF1 and CF2 are identical, we know if we satisfy one of the equations we are guaranteed to satisfy the other. Thus, eliminate the equation labeled CF2, which leaves three equations in three unknowns. The solution to the remaining three equations is NA = 3 NB = 2 NC = 1 . Clearly, both formulations provide the same set of numerical answers. Solutions to Interpreting Bond Yields
1. Part a. To calculate the yields on the 3 bonds, solve: 550 100 = y 2 yA = 85.24% (e A ) 100 = 100 = 225 yB = 81.09% (eyB ) 450 yC = 150.41% . (eyC ) The market value, V , for each bond is: 550 VA = 0.1 2 VA = 450.30 (e ) VB = VC = 225 VB = 203.59 (e0.1 ) 450 VC = 407.18 . (e0.1 ) Thus, Bond A is the most undervalued bond even though Bond C has the highest yield. Part b. Let portfolio D consist of Bond A and C. Let portfolio E consist of Bonds B and C. To calculate the yield on each portfolio: 550 450 200 = y + y 2 (e D ) (e D ) yD = 114.07% . 200 = 675 (eyE ) yE = 121.64% . Page 1346 Chapter 29: Solutions To Questions, Bond Valuation Using Synthetics Page 1345 Chapter 29: Solutions To Questions, Interpreting Bond Yields Adventures in Debentures Adventures in Debentures The market value of each portfolio is: 550 450 VD = 0.1 + 0.1 2 = 857.48 (e ) (e ) = VA + VC = 450.30 + 407.18 . VE = 675 = 610.77 (e0.1 ) = VB + VC = 203.59 + 407.18 . 4.) Calculate the "annuity yield," at (for t = 1, 2, . . . , T ). That is, find at such that the left hand side and the right hand side of the following equation are equal: At = 1 1 1 + + . + (1 + at ) (1 + at )2 (1 + at )t Note that the yield on portfolio E is higher than the yield on portfolio D. However, portfolio D is more undervalued than portfolio E. Clearly, the net preset value (NPV) is additive. That is, the NPV of portfolio D is equal to the NPV of Bond A plus the NPV of Bond B. Similarly, the NPV of portfolio E is equal to the NPV of Bond B plus the NPV of Bond C. However, yield on portfolio D does not equal a simple weighted average of the yields on Bonds A and B; similarly, the yield on portfolio E does not equal a simple weighted average of the yields on Bonds B and C. (There are two obvious methods to use to weight the yields on the individual bonds to get the yields on the portfolio. You could try equal weighting (since you hold 1 unit of each bond and/or since each bond costs $100. Alternatively, you could valueweight using the true values of each bond. Neither weighting scheme will give you the yield on the portfolio using the yield on the individual bonds.) 5.) Calculate the price, Pt , of a constant coupon bond with coupon equal to K and maturity t, and face value of $100. Pt = Kd1 + Kd2 + + (K + 100)dt . 2. To create plots similar to those in the Schaefer article, the following steps should be followed: 1.) Specify values for r1 , r2 , . . ., rT . These represent the term structure of annually compounded rates based on zeros. 2.) Calculate the discount function (d1 , d2 , . . ., dT ) where dt = 1 . (1 + rt )t 6.) Calculate the yield to maturity, y, on this constant coupon bond. That is, solve for y: Pt = K K + 100 K + + . + (1 + y) (1 + y)2 (1 + y)T 7.) Draw rt , at , and yt as a function of t. The attached figures illustrate my attempt to create graphs similar to those in the Schaefer article. In the graphs that follow the constant coupon bond has a coupon yield of 5%.
Chapter 29: Solutions To Questions, Interpreting Bond Yields Page 1348 3.) Calculate the price of an annuity, At (for t = 1, 2, . . . , T ), that pays $1 each year for t years. That is, At = d1 + d2 + + dt .
Chapter 29: Solutions To Questions, Interpreting Bond Yields Page 1347 Adventures in Debentures Adventures in Debentures Yield Curves 12.00% 10.00% 8.00% Y i e l d s 6.00% 4.00% 2.00% 0.00% 1 5 9 13 17 21 25 29 33 37 Maturity (in years) Yield Curves 25.00%
.. ... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... . . . . . . . . . . . . . . . . . . . . . . .... . . . . . . . . . .... . . .... . . . . .... . . . .... .... ... ... ... ... ... ... ... ... ... .... .... .... .... .... .... .... .... .... . . . .... ... ... ... .... . .... .... .... . . .... ... ... ... .... ... .... ... ... ... . .... .... .... . . . .... ... ... .... .... .... ..... . . . .. ... ... .... .... ... . .. .... . ... .. .... .... . . .. .. .. .. ... .. .... ..... ... . ... . .... . ... . .... .... .... ... ... .... ... .... . .... .... .... ... .... .... ............... ............... ... . ... ... .... . ..... . ...... ...... .. .. .... .... ... ... ... .. . . . . . .............................................. ............................................... ......... ....... ......... ....... ....... ....... ...... ...... ...... ...... ..... ..... ..... ..... .... ... . ..... ..... .... . . . . . . . . . . ........ .... .... . . . . . . . ... . .... .... . ... . . ... ... ....... ... ... ... ... ... . .... .... .... .... ....... ..... ..... .... .... .... . . . . .... .... ... ... ... ... ... .... ... ... . . . .... ... . . . . . ... . .... ... .. .... . .. . .. . . ... ... . . .... . ... .. .... .... . ... ... . . ... .. .. ... ... .. . . ... ... ... ... . . . ... ... ... ... ... . .... ... . . ... . ... ... . ... .... ... ... ... ... .... ... ... . . ... ... ... .... ... ... ... . . ... ... .... ... ... ... . . .. .. . .. ... .... .. . ... ... . ... ... ... . ... .. ... . . ... ... . .. .. . ... ... . . ... ... .. ... . .... .. .. . .. ... .. . ... .. .. .. ... .. .... .. . .. . .. ... ...... .. . .. ... ... 3. We need to construct the rates for zeros from the par yield curve. Since the par yields are quoted on an annually compounded basis, it will be more direct to compute the rates for zero coupon bonds on an annually compounded basis. We need r1 , r2 , and r3 . Since a 1 year constant coupon bond is just a one year zero, we know r1 = 10%. Now we know from the par yield curve 100 = 115 15 + (1.1) (1 + r2 )2 r2 15.39% . = ............... ............... .... .... ... ... ... .. . . . . . Yields on Zeroes Annuity Yields Yields on Coupon (5%) Bonds Similarly, from the par yield curve 100 = 20 120 20 + + (1.1) (1.1539)2 (1 + r3 )3 r3 21.56% . = Now, the current value of the bond is just 10 110 10 110 10 10 + = + = 77.84 . + + (1 + r1 ) (1 + r2 )2 (1 + r3 )3 (1.1) (1.1539)2 (1.2156)3 20.00% Y 15.00% i e l d s 10.00% 4. For all parts of this question, the Schaefer article refers to the 1977 article titled "The Problem with Redemption Yields" in the Financial Analysts Journal. Part a. True. (The Schaefer article makes this point.) Part b. False. (The Schaefer article only established the bounds. Even if yield falls within the bounds, arbitrage may exist.) Part c. True. (This is an example of a "crossing point" in yield curves from the Schaefer article.) Part d. True. (See page 64 of Schaefer article.)
Chapter 29: Solutions To Questions, Interpreting Bond Yields Page 1350 5.00% Yields on Zeroes Annuity Yields Yields on Coupon (5%) Bonds 0.00% 1 5 9 13 17 21 25 29 33 37 Maturity (in years)
Chapter 29: Solutions To Questions, Interpreting Bond Yields Page 1349 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Part e. True. (Consider a flat term structure of interest rates.) Part f. False. (The yields on all zero coupon bonds with the same maturity must be equal for any face value.) Part g. False. (See pages 6364 of Schaefer article.) Part h. True. (A par bond is just a special case of a constant coupon bond.) Part i. False. (The Schaefer article is about constant coupon bonds. This is not a constant coupon bond. Consider an example where r1 = 10.00% and r2 = 10.25%.) Part j. True. (Example: r1 = 10%, r2 = 11%, r3 = 12%, r4 = 13%, r5 = 12%, r6 = 11%, r7 = 10%, r8 = 9%, r9 = 8%, r10 = 7.5%, r11 = 8%, r12 = 9%, r13 = 10%.) Solutions to Bond Values and the Passage of Time
1. Part a. .952569  .937022 = .077735 .7 = .2 Part b. coup can be calculated by recognizing that the synthetic alternative to a coupon is a portfolio of zeroes. Further, the of a portfolio is a weighted sum of the 's for each component. coup = 150 .5 + 150 1.0 + 150 1.5 + 1,150 2.0 = (150 .078407) + (150 .076705) + (150 .074908) + (1,150 .073029) = 118.49 Part c. is measured as a capital gain due to the passage of time. The units of are gains per year. Therefore, approx. cap. gain for 1 month = 118.49/12 = 9.87 . Part d. portfolio = (2,000 .069081) + (4,500 .073029)  (1,200 .078407)  (700 .076705) = 319.01 Part e. approx. capital gain = 319.01 2 12 = $53.17 Chapter 29: Solutions To Questions, Interpreting Bond Yields Page 1351 Chapter 29: Solutions To Questions, Bond Values and the Passage of Time Page 1352 Adventures in Debentures Adventures in Debentures 100 Long Forward To achieve the goal discussed in the first paragraph, we want to maximize of the net equity. To maximize , you do not necessarily want to buy the security with the highest and sell the security with the smallest . Such an approach neglects the cost of the security. To maximize subject to a wealth constraint, you want to buy the security with the highest (/PRICE) and sell the security with the lowest (/PRICE). Part b. Even when the net equity is zero, the approach of part a remains applicable. Part d Dollars Invested 10,000 40,000 100,000 100,000 0 100,000 Total 30,000 40,000 40,000 40,000 Part c Dollars Invested 50,000 50,000  By using forward contracts, you can increase the for your net equity position. Part e. The theta values for each of the four portfolios in parts a through d are summarized at the bottom of the following table. As expected a exceeds b , for the market value of the assets in part b is less than the market value of the assets in part a. As expected a exceeds c , for the solution in part c had more restrictions imposed on it. It is impossible for c to be larger than a ; at best, the thetas of the 2 positions could be equal if the restriction in part c was not binding. As expected d exceeds c , for position d may have more effective leverage than position c.
Part a Dollars Invested 150,000 50,000  /PRICE 8.07% 8.67% 8.55% 9.54% 7.79% 7.40% ID Chapter 29: Solutions To Questions, Bond Values and the Passage of Time Page 1353 Chapter 29: Solutions To Questions, Bond Values and the Passage of Time G D A C B E F  Net Equity 0 0 0 0 = $10,614.71 Units Invested 16.56 537.35  0 0 0 0 Net Equity 0 0 0 0 Part d. Even though (/PRICE) is not defined for a forward contract, you want to go long a forward contract if its is positive. (You will short the forward contract if its is negative.) Part b = $1,070.03 Part c. With the diversification constraint, you want to hold securities by the descending order of (/PRICE). You want to short securities by the ascending order of (/PRICE). You want the maximum leverage as long as you can take the borrowed money and invest in a security with a (/PRICE) higher than the security that you short. We expect Net Equity in part a to be greater than Net Equity in part c. Dollars Invested 30,000 40,000 40,000 40,000 10,000 40,000  Units Invested 5.52 537.35  0 0 0 0 Net Equity = $9,383.80 Units Invested 2,915.45 61.02 37.34 4.42 115.96 429.88  Net 0 Equity = $10,287.80 Units Invested 2,915.45 61.02 37.34 4.42 115.96 429.88 2. Part a. See the following table which summarizes the answer for all parts to this question. Page 1354 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Solutions to Forward Contracts
1. Part a. 0 r 12 = 20 r02  0 r01 = (2 7.1531%)  6.1875% = 8.1187%. To create a zero coupon yield curve where Long Forward is negative, it is easier to work with the discount function and then transform the discount function into a zero coupon yield curve. A discount function is economically reasonable as long as it is between 1 and 0; furthermore, the discount function should have a negative slope at all maturities. A negative forward rate is not economically reasonable. The following table provides an example where the forward contract has a negative theta. This discount function is economically reasonable. Thus, the forward price = 1,000 e1.081187 = 922.02. Part b. A synthetic alternative to this forward contract involves shorting 1year paper and going long 2year paper. Specifically, N1 = 922.02 and N2 = 1,000.00. Now we need thetas for 1year and 2year paper: 1 = d.75  d1.25 e.75.060834  e1.25.066705 .955400  .920000 = = = .070800 .5 .5 .5 Maturity 0.75 1.00 1.25 1.75 2.00 2.25 e Current Discount .955400
.061875 New Discount .970000 .940000 .920000 .884300 .866700 .844999 New Zero Coupon Yield 4.0612% 6.1875% 6.6705% 7.0262% 7.1531% 7.4853% = .940000 .920000 .884300 e2.071531 = .866700 .844999 d  d2.25 e1.75.070262  e2.25.074853 .884300  .844999 2 = 1.75 = = = .078602 .5 .5 .5 Thus, Long
Forward By just setting d.75 = .970000, we now achieve a negative theta: 1 = Long
Forward = (922.02 .070800) + (1,000 .078602) = 13.32 . .970000  .9220000 = .1 .5 = (922.02 .1) + (1,000 .078602) = 13.60. Part c. This new discount function is economically reasonable. This new discount function does not change the forward rate of 8.1187% calculated in part a above. Short Part d.
Chapter 29: Solutions To Questions, Forward Contracts Page 1355 Chapter 29: Solutions To Questions, Forward Contracts Page 1356
Forward = (922.02 .070800) + (1,000 .078602) = 13.32 . 2. Adventures in Debentures Adventures in Debentures By the definition of a par bond: 0.10(100) 100 100 = + r (er1 ) (e 1 ) 100 = r1 = 9.53% 4. Part a. B = 10,000 e.10(30/365) = 9,918.15 r2 = 14.32% . Part b. 0 = B e.20(5/365) + 10,000 e.10(30/365) B = 9,945.36 0 r 1,2 = 19.11% . 5. 0.15(100) 0.15(100) 100 + r 2 + (er1 ) (er2 )2 (e 2 ) To find the relevant forward rate, note that (e0 r0,1 )(e0 r1,2 ) = (e0 r0,2 )2 or 0.0953 + 0 r1,2 = 2(0.1432) 3. The following time line describes the relevant cash flows:
7 days from today To set up the forward transaction South must both borrow and lend. Since South is receiving 30,000,000 dollars in one year, he can borrow 30,000,00/e.14 = 26,080,747.06 dollars this year and have exactly enough to repay that loan when he receives the money from Tola. Note that we use here the borrowing rate of 14% which South faces. South can then take the money he receives when he borrows and invest it in a three year CD. At 13% interest this grows to be 38,520,762.50 dollars. Thus South takes the 30,000,000 that he will received in one year and converts it into 38,520,762.50 dollars to be received in 3 three years. The forward lending rate for this loan is
1 2 1 2 3 4 constant coupon bond (A) Forward Contract (B) 998.20 100 100 100 1100 ln 38,520,762.50 = 0.125 . 30,000,000 0 0 8,203.24 0 10,000 The calculation above does not include the receipt of the toaster. We now redo the calculation, this time accounting for the fact that South will receive a toaster when he puts well over 26,000,000 dollars on deposit with the bank. We will assume that the toaster is worth 45 dollars (one that can toast four slices of bread at a time might cost this amount). If South sells the toaster and invests this along with the 26,080,747.06 dollars he receives when he borrows, then he will have a total of 26,080,792.06 dollars growing at the rate of 13% over 3 years. This means he will have at the end of three years 38,520,828.96 dollars. The forward rate is now equal to
1 2 Annuity (C) 2 year zero coupon bond (D) 3.1762 1 1 1 1 3,404.15 0 4,101.62 0 0 Clearly, to replicate the forward contract, you need to undertake the following strategy: Issue 1,000 units of C (that is, NC = 1,000), and buy 10 units of A (i.e., NA = 10). ln 38,520,828.96 = 0.125000862 . 30,000,000
Page 1357 Chapter 29: Solutions To Questions, Forward Contracts Chapter 29: Solutions To Questions, Forward Contracts Page 1358 Adventures in Debentures Adventures in Debentures Now issue 2 units of D (that is, ND = 2). This will replicate forward, and it costs: (10 998.20)  (1,000 3.1762)  (2 3,404.15) = 2.50 . The explicit forward has 0 value. Thus, the explicit forward is overvalued by market, and we should buy the synthetic forward and short the explicit forward. Such a transaction will generate a profit of $2.50. There is an alternative, but equivalent, solution approach to this problem. The prior (and perhaps more intuitive) solution tries to find a synthetic and then compares the cost of the synthetic to the actual security. If the costs of the actual and the synthetic are different, then arbitrage is possible by buying the cheap and selling the expensive. The second method is similar, but more direct; this second approach is described below. It looks for an arbitrage strategy without doing the intermediate step of constructing a synthetic. We begin by formulating an algebraic statement of the desired arbitrage position. We want the cash inflow in year 0 (CF0) to be +2.50. We want the cash flow in 7 days or years 1, 2, 3, and 4 (that is, CF7, CF1, CF2, CF3, and CF4) to be zero. If we could find such a position, we have found a classic arbitrage. Thus, we want: 0NB  3.1762NC  3,404.15ND = 2.50 CF 7 : 998.20NA + 0NB + 1NC + 0ND = 0 CF 1 : 100NA + 1NC + 4,101.62ND = 0 CF 2 : 100NA  8,203.24NB + 0NB + 1NC + 0ND = 0 CF 3 : 100NA + 10,000NB + 1NC + 0ND = 0 CF 4 : 1,100NA + Since the equations for CF1 and CF3 are identical, we know if we satisfy one of the equations we are guaranteed to satisfy the other. Thus, eliminate the equation labeled CF3, which leaves four equations in four unknowns. The solution to the remaining four equations is NA = 10 NB = 1 NC = 1000 ND = 2 . Clearly, both formulations provide the same set of numerical answers. Now, there is an obvious synthetic which you can buy today, and in 180 days it will be worth the above amount. This synthetic consists of 3 components: Today issue bonds that mature in 180 days with face value = 50,000 + 25,000 f . Today buy 1,000 units of 90 + 180 = 270 days till maturity. Each bond should have face value = $100. Today issue 250 units of 360 + 180 = 540 days till maturity. Each bond should have face value = $100. When these three components are held for 180 days, they will have value in total equal to the payoff of the YCS forward contract. Part b. Now we want to find f such that the above synthetic has value = 0 as of today: (50,000 + 25,000 f )d180  1,000(100 d270 ) + 250(100 d540 ) = 0 (50,000 + 25,000 f )(.97482)  (1,000 100 .96175) + (250 100 .91119) = 0 f 1.01164 . = 6. Part a. First, rewrite the payoff to a YCS forward contract in terms of prices, not yields: 100  B360 360 100  B90  f 25,000(1 + y360  y90  f ) = 25,000 1 + 100 90 100 = 50,000  25,000 f + 1,000 B90  250 B360 .
Chapter 29: Solutions To Questions, Forward Contracts Page 1359 You were not asked the following question, but given the above solution it is an easy extension. You may want to return to the following question after we study dollar duration and convexity in the upcoming chapters. Based on the information in part a above, what is the dollar duration (or delta) and gamma for this YCS forward contract from the perspective of the long side of the forward contract? To find the and of the YCS forward contract, we can find the and of the
Chapter 29: Solutions To Questions, Forward Contracts Page 1360 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons above synthetic. = (180/365) P V180 [50,000  (25,000 1.01164)] 100 (270/365) P V270 (1,000 100) (540/365) P V540 (250 100) + + = 12.47 100 100 (180/365)2 P V180 [50,000  (25,000 1.01164)] 10,000 (270/365)2 P V270 (1,000 100) (540/365)2 P V540 (250 100) + + = 1.51 . 10,000 100 Solutions to Dollar Delta 1: Risk Measurement
1. We know a T period constant coupon bond can be viewed as a portfolio of a T period annuity and a T period zero coupon bond. We also know the delta of a portfolio is the sum of the deltas of the components of the portfolio. Thus, $coupon = $annuity + $zero . = We are given $coupon = 8.2604 and we can compute $zero =
100 (1.20)30 30 100 = 0.1264 . Thus, $annuity = 8.2604  0.1264 = 8.1340 . 2. Part a. When r1 = 8%, the problem indicates the dollar value of the net equity is $39.87. When r1 = 6%, then with a uniform shift r2 = 6.5% and r3 = 8% Vne = 20 e.06  80 e.0652 + 120 e.083 42.98 = When r1 = 11%, then with a uniform shift r2 = 11.5% an r3 = 13% Vne = 20 e.11  80 e.1152 + 120 e.133 35.60 = Part b. When r1 = 9%, we know that the rest of the term structure shifts up by a uniform amount. Thus, r2 = 9.5% and r3 = 11%. Chapter 29: Solutions To Questions, Forward Contracts Page 1361 Chapter 29: Solutions To Questions, Dollar Delta 1: Risk Measurement Page 1362 Adventures in Debentures Adventures in Debentures Now, we need to solve for r4 : 90.00 = 10 e .09 if a1 = 11% and a2 = 16%, then: + 10 e
.0952 + 10 e .113 + 110 e r4 4 r4 13.00% = 100 e.11 + 110 e.162 + 110 e.16 84.03 = if a1 = 9% and a2 = 14%, then: 3. To make this problem easier, a little algebra will help. Let At = the market price of a t period annuity that pays $1 for each year until year t. Hence, the connection between At and dt is: A1 = 1d1 A2 = 1d1 + 1d2 A3 = 1d1 + 1d2 + 1d3 etc. d1 = A1 d2 = A2  A1 d3 = A3  A2 etc. 100 e.09 + 110 e.142 + 110 e.14 = 87.37 Value ($)
.. ... . . $87.37 $85.68 $84.03 . .... ... . . Let at = the continuously compounded yield to maturity on a tperiod annuity. That is, A1 = ea1 A2 = e
a2 a1 +e a2 2 9% 10% 11% A3 = ea3 + ea3 2 + ea3 3 . Now, let's value the 2 year coupon bond that pays 10 in period 1 and $110 in period 2: B2yr coup = 10 d1 + 110 d2 4. Part a. r1 r1 4.87902% r6 6.76586% r6 7.00% 6.75% 7.25% B6yr
1000/ (1.07)6 1000/ (1.0675)6 1000/ (1.0725)6 by substitution: = 10 A1 + 110(A2  A1 ) = 10 ea1 + 110(ea2 + ea2 2  ea1 ) = 100 ea1 + 110(ea2 + ea2 2 ) . Now we are ready to value the 2 yr. coupon bond under different scenarios about the annuity yield curve: if a1 = 10% and a2 = 15%, then 100 e.10 + 110 e.152 + 110 e.15 85.68 =
Chapter 29: Solutions To Questions, Dollar Delta 1: Risk Measurement Page 1363 5.00% 4.75% 5.25% = 666.34 = 675.76 = 657.08 Note: r1 = ln(1 + r1 ). Since r1 in the first row matches the value of r1 in the current term structure within the question, I can use the current term structure to provide r6 in the first row. Given r6 , I know r6 = er6  1. To compute rows 2 and 3, I do not need r1 or r6 . I know all shifts in the annually compounded yields are uniform. Chapter 29: Solutions To Questions, Dollar Delta 1: Risk Measurement Page 1364 Adventures in Debentures Adventures in Debentures . . ... ... . . . . . . Part d. The time profile can be created without any assumptions about potential shifts in the term structure. Thus, the time profile is completely unaffected. .. . ....... ...... . . 675.76 Value of 6yr 666.34 657.08 4.75% 5.00% r1 5.25% Part b. Approximate DeltaLike Measure for Uniform Shifts in Annually Compounded Yields = 675.76  657.08 = 37.36 4.75  5.25 This is approximately minus the slope of the performance profile when r1 = 5.00%. Part c. $ = 6 666.34 6 P V (1000) = = 39.98 > 37.36 . 100 100 (It is also acceptable to compute $ by approximating minus the slope of the performance profile when you assume uniform shifts in the continuously compounded yields.) To understand why $ is larger here than in b above, consider the change in the annually compounded yield when the continuously compounded yield changes by 100 basis points: If r6 = 6.76586% and it decreases to 5.76586%, then r6 moves from 7.0% to 5.93533% (a drop of 106.47 basis points). If r6 = 6.76586% and it increases to 7.76586%, then r6 moves from 7.0% to 8.07536% (an increase of 107.54 basis points). Thus, a 100 basis point change in the continuously compounded yield implies a change of more than 100 basis points in the annually compounded yields. However, the deltalike calculation in part b above was for only a 100 basis point change in the annually compounded yields, so part b will produce a smaller number than part c.
Chapter 29: Solutions To Questions, Dollar Delta 1: Risk Measurement Page 1365 Chapter 29: Solutions To Questions, Dollar Delta 1: Risk Measurement Page 1366 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures the following securities: Solutions to Dollar Delta 2: Risk Management
1. Part a. In general, the market value of any asset can be calculated using:
T 1 = 1d1 = .009524 100 B = 5 15 t=1 (1+r )t t t+ 100 (1+r5 )5 5 3d 3 = 3 = .024489 100
10 t=1 4.9553 =
20 t=1 100 (1+r10 )10 100
5 (1+rt )t t + 20 t A = 5.8480 = 10 (1+rt )t C = 3.6613 = D =
20 t=1 100 (1+r20 )20 + 10 100 100 5t 100 = 10.50 Value =
t=1 Kt , (1 + rt )t The value of the net equity, Vne , is: Vne = NC BC + ND BD + N1 d1 + N3 d3 + NB BB = (2,400 44.3391) + (2,000 100)  (50,000 .9524)  (60,000 .8163)  (500 125.6477) = 146,993.15 . The delta of the net equity, ne , is: ne = NC C + ND D + N1 1 + N3 3 + NB B = 25,363.9483 and the delta of any security is given by:
T t=1 Kt (1+rt )t = T t=1 t PV(Kt ) = 100 T t=1 t Kt etrt t 100 = 100 Using the first equation, we can calculate the current market values for the following securities:
5 d1 = 1 /1 + r1 = .9524 BB =
t=1 100 15 + (1 + rt )t (1 + r5 )5 100 5 + (1 + rt )t (1 + r20 )20 5 dt The following graph provides the performance profile of total assets (the top curve), total liabilities (the bottom curve), and net equity (the middle curve). The scale of the vertical axis was determined by the second graph, which appears later. Part b. Given the value of the net equity is 146,993.15 and the price of a 2 year zero per $1 face value, d2 , is .8900, one could purchase 165,160.84 units of a 2 year zero with the net equity (i.e., 165,160.84 = 146,993.15/.8900). The delta from this number of units of a 2 year zero is: 146,993.15 .8900 Note:
(2.8900) 100 125.65 =
20 d3 = 1 (1 + r3 )3 = .8163 BC =
t=1 44.34 =
10 BA =
t=1 100 10 92.42 + = (1 + rt )t (1 + r10 )10 20 BD =
t=1 dt = 5 20 = 100 2 .8900 100 = 2,939.8630 . Using the general equation for given above, we can calculate the delta values for
Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1367 is the delta for each unit of a 2 year zero. Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1368 Adventures in Debentures Adventures in Debentures Thus, to achieve the desired interest rate exposure, we need to solve the following 2 equations: Vne = 146,993.15 = NC BC + ND BD + NB BB + NA BA + NB BB = (2,400 44.3391) + (2,000 100)  (500 125.6477) + 92.4167 NA + 125.6477 NB ne = 2,939.8630 = NC C + ND D + NB B + NA A + NB B = (2,400 3.661304) + (2,000 10.50)  (500 4.955313) + (5.848012 NA ) + (4.955313 NB ) We can simplify the last 2 equations to find: 92.4167 NA + 125.6477 NB = 96,596.9202 5.848012 NA + 4.955313 NB = 24,369.612055 , where the primes on the NA and NB indicate that these 2 assets are being used to achieve our desired delta. The values of NA and NB satisfying the above 2 equations are: NB = 6,094.8056 NA = 9,331.5945 D o l l a r V a l u e s 1,500,000 Performance ProfileOriginal Balance Sheet 1,000,000 500,000 . .. .. .. .. .. .. ... .. .. .... .... 2,400 units of C and 2,000 units of D Net Equity . .. .. .. .. .. ... .. .. . .... .. .. . .. . ....
. 0 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 ... ...
.... . .... . .. . ... ... ... ... ... ... ... ... ... .. . Part c. The following graph provides the performance profile after the balance sheet was restructured. The top curve is for total assets, the bottom curve is for total liabilities, and the middle curve is net equity. The next graph provides a performance profile when there are no liabilities and 165,160.84 units of the 2 year zero are held. The contrast between the first two figures is striking, and the similarity between the last 2 figures is as we expect. 500,000 50,000 units of 1 year zero and 60,000 units of a 3 year zero and 500 units of B 1,000,000 1,500,000 Continuously Compounded 1 Year Rate Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1369 Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1370 Adventures in Debentures Adventures in Debentures Performance ProfileRestructured Balance Sheet 1,500,000 1,000,000 2,400 units of C and 2,000 units of D and 6,094.81 units of B ... ... . 1,000,000 1,500,000 Performance Profile165,160 Units of 2 Year Zero .. ... ... ... ... . ... .... . .. .. ... . .... 500,000 500,000 D o l l a r V a l u e s
.. ... ... . .. .... .... .... .... Net Equity . ... ... 0 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 D o l l a r V a l u e s 0 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 500,000 ... ... ... .. 500,000 1,000,000 1,500,000 ..................... 500 units of B and 9,331.59 units of A 1,000,000 1,500,000 Continuously Compounded 1 Year Rate Continuously Compounded 1 Year Rate Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1371 Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1372 Adventures in Debentures Adventures in Debentures 2. Part a. We have to find the appropriate forward price, X, such that the forward contract has zero present value when it was negotiated. That is, 0 = X e2r2 + 70 e3r3 + 70 e4r4 + 1,070 e5r5 . Solve for f : Forward Price = 72.5091 . Part b. 47,000 = (10,000 .5322) + (.8261 NLeap ) + 2.3099N1yr zero 43,000 = (10,000 4.2515) + (5.2081 NLeap ) + 0.0633N1yr zero . Solving the 2 above equations in the 2 unknowns implies: Now solve for spot rates based on forward rates. Since forward rates are continuously compounded, we know 0 r 0t Thus, 0 r 02 0 r 03 0 r 04 0 r 05 = + .06) = .05 = 1/3(.04 + .06 + .11) = .07 = 1/4(.04 + .06 + .11 + .15) = .09 = 1/5(.04 + .06 + .11 + .15 + .14) = .10 . 4. Part a.
1/ (.04 2 = 1/t(0 r01 + 0 r12 + 0 r23 + . . . + 0 rt1,t ) . N1yr zero = 18,088.52 NLeap = 126.726 . Now, return to the first equation, and solving for X: X = 70 e(2.053.07) + 70 e(2.054.09) + 1,070 e(2.055.10) 833.92 . = Benchmark = 0.28 + 31.34 + 44.93 + 1.72 + 0.93 = 79.20 Part b. $ =
1 [2 100 833.92 e2.05 + 3 70 e3.07 + 4 70 e4.09
5.10 Value of Benchmark VBenchmark = 10.29 + 655.53 + 1,071.20 + 86.24 + 93.05 = 1,916.31 Benchmark = Benchmark 79.20 = = 4.13% VBenchmark 1,916.31 + 5 1,070 e 21 . = Part c. $5yr zero = 52,078.75 e5.10 100 63. = Part b. Since each unit of the benchmark costs 1,916.31, you can buy 50,000,000 = 26,091.81 units 1,916.31 with $50,000,000. If your target is 2.00% with a $100,000,000 investment, then your target = .02 100,000,000 = 2,000,000. To structure your portfolio, you need to solve the following 2 equations: 10.29 NA + 655.53 NB + 50,000,000 = 100,000,000 .28 NA + 31.34 NB + 79.20 26,091.81 = 2,000,000
Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management To obtain $net equity = 0, we need 0 = 63 + 21 Nforward Nforward = 3 . Hence, you need to short 3 forward contracts. 3. Part a. Find the Forward Price, f , such that: 0 = f e.11164 + 15 e.10048 + 15 e.089212 + 15 e.078016 + 115 e.066820 .
Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1373 NA = 11,591,850.06 NB = 105,685.69
Page 1374 Adventures in Debentures Adventures in Debentures ID A B C D E F Units Invested 26,091.81 + 11,591,850.06 = 11,617,941.87 26,091.81  105,685.69 = 79,593.88 26,091.81 0 26,091.81 26,091.81 ID A B C D E F G Units Invested 52,183.62 52,183.62 52,183.62 52,183.62 52,183.62 52,183.62 76,203.74 Part c. Price Change (Change in Rates) = Return = Price Change (Change in Rates) = Price Price = (Change in Rates). (i.e., a short position in the forward contracts) Part e. To create a perfect substitute for H, buy 11 units of E and 1 unit of F. If we go long 1 unit of H and short 11 units of E and short 1 unit of F, we profit by If change in rates = 50 basis points, the (Change in Rates) = .5. If Benchmark = 4.13% and Actual = 2.00%, then, (Return on Benchmark) 4.13% .5 = 2.065% = (Return on Actual) = 2.00% .5 = 1.00% Thus, if the term structure increases by 50 basis points, the actual portfolio will have an approximate return of 1.00%. Similarly, if the term structure increases by 50 basis points, the benchmark portfolio will have an approximate return equal to 2.065%. Thus, the portfolio will outperform the benchmark: 1.00%  (2.065%) = +1.065%. (11 86.24) + (1 93.05)  1,000 = 41.69. To maximize our profits, we need to hold the maximum arbitrage position which is 3000 units of H, short 3000 units of F, and short 11 3000 units of E. This will create an immediate profit: 41.69 3000 = 125,070 . with no future cash flows. Thus, you need to structure your portfolio to manage a wealth of 100,000,000 + 125,070, and your target remains 2.00%. (However, your target delta is now .02 100,125,070, which reflects the immediate increase in your wealth due to the arbitrage.) The units to invest in the benchmark will remain as in part b above, so solve: Part d. Since each unit of the benchmark costs 1,916.31, you can buy 100,000,000 = 52,183.62 units 1,916.31 with $100,000,000. If your target is 2.0%, this implies your target is .02 100,000,000 = 2,000,000. With forward contracts, we need not worry about a value constraint. Thus, to structure the portfolio, we only need to solve: 27.99 NG + (52,183.62 79.20) = 2,000,000 NG = 76,203.74
Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1375 10.29 NA + 655.53 NB + 50,000,000 = 100,000,000 + 125,070 0.28 NA + 31.34 NB + (79.20 26,091.81) = .02(100,000,000 + 125,070) NA = 11,608,259.64 and NB = 105,752.48.
Page 1376 Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons ID A B C D E F G H Units for Benchmark 26,091.81 26,091.81 26,091.81 0 26,091.81 26,091.81 0 0 Units for Arbitrage 0 0 0 0 33,000 3,000 0 3,000 Units to Obtain Target Omega 11,608,259.64 105,752.48 0 0 0 0 0 0 Total Units 11,634,351.45 79,660.67 26,091.81 0 6,908.19 23,091.81 0 3,000.00 Solutions to Dollar Gamma
1. Part a. True. approximate price change = 550(.20)  1/2(20)(.2)2 = 110.40 Part b. False. only measures the curvature of the performance profile at one particular point. Since the present value rule is not a linear expression, it is hard to imagine how the performance profile could be a straight line. Part c. False. is minus the slope of the delta profile. Part d. False. measures the change in delta for a 100 bp move, not a 1 bp move. Thus, predicts the new delta to be 750  150 = 748.5. 100 Part e. True. If the short delivers a par bond, then the value of the forward contract is always 0, no matter what the level of interest rates. That is, the discounted value of zero is always zero. Part f. True. From the perspective of the party who is long, they are obligated to pay $110 for something worth $100. So the long always has a liability of $10. However, the P V of this $10 will vary as interest rates change. This is just like any other liability associated with a zero coupon bond. Hence, the must be negative. Part g. True. With a uniform shift, the particular maturity used as indicator of term structure for the horizontal axis cannot affect the curvature of the performance profile. Since the shifts are uniform, the change in the value on the vertical axis for a given change in the interest rate on a zero coupon bond (the horizontal axis) will be the same for any maturity selected on the zero coupon bond. Part h.
Chapter 29: Solutions To Questions, Dollar Delta 2: Risk Management Page 1377 False. If the term structure is not flat, then annually compounded yields
Page 1378 Chapter 29: Solutions To Questions, Dollar Gamma Adventures in Debentures Adventures in Debentures will not shift uniformly even if continuously compounded yields do. As a result, the curvature of the performance profile will not be the same when comparing a profile using continuously compounded yields on the horizontal axis versus a profile for the same security using annually compounded yields on the horizontal axis. Part i. False. In general, we expect a portfolio of fixed income securities to have a nonzero gamma. Since a forward contract can be viewed as a synthetic portfolio of bonds, we expect its gamma to be nonzero as well. If interest rates shift up by 300 basis points, then net equity equals approximately: 90,355.82 + 189,487 = 279,842.82. If interest rates shift down by 300 basis points, then net equity equals approximately: 189,487  90,355.82 = 99,131.18. Approx. Perf. Profile using :
Net Equity
. . ... . . . . . . . . . .. .. ... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... ... ... ... ... . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... Approx. Perf. Profile using and :
Net Equity $189,487.00
. . ... . . . . . . . . . .. . .... .... .... ... ... ... ... ... ... ... ... ... ... .. .. ... . ... ... ... ... ... .. ... ... ... ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. $279,842.82 .. . . $266,016.36...... . 2. Part a. Approximate 5% =  49.588349.9866 9.9575 = 5.024.98 (That is, we are approximating the slope of the performance profile at 5% by measuring the price impact of a 4 basis point change.) Part b. Approximate 5.04 =  49.193249.5883 9.8775 = 5.065.02
5.04 5.00 Part c. Approximate 5.02 =  5.045.00 =  9.87759.9575 = 2 5.045.00 $189,487.00 99,131.18 . . .... .. . 0 r 0,1 85,304.73 . . .... .. . 0 r 0,1 11% 14% 17% 11% 14% 17% Part b. (change in equity) = $ (change in r) + 1 (change in r)2 2 change in equity for 300bp increase = 90,355.82 + 1 (3,072.545)9 = 76,529.368 2 = 90,355.82 + 1 (3,072.545)(9) = 104,182.27 2 3. Net Equity = 1,000 2 + 250 4  1,300 7 = (1,000 15.4210) + (250 25.7615)  (1,300 39.9846) = 30,118.605 NE = 1,000 2 + 250 4  1,300 7 = (1,000 .3084) + (250 1.0305)  (1,300 2.7989) = 3,072.545 VNet Equity = (1,000 771.05) + (250 644.04)  (1,300 571.21) = $189,487 change in equity for 300bp decrease (net equity if r1 = 17%) = 189,487.00 + 76,529.368 = $266,016.36 (net equity if r1 = 11%) = 189,487.00  104,182.27 = $85,304.73 Part a. (change in net equity) = NE (change in r) = +30,118.605 3 +90,355.82 =
Chapter 29: Solutions To Questions, Dollar Gamma Page 1379 Chapter 29: Solutions To Questions, Dollar Gamma Page 1380 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Solutions to Delta, Gamma, and Theta
1. Part a. Find the forward price, F , such that the following is true: 0 = F e2.083194 + 10,000 e6.088750 $ = F = 6,934.25 Notice that the current balance sheet already achieves the desired delta. Further, the combination of forward contracts in part b provides a zero delta. Thus, using the combination of forward contracts in part b provides a strategy which has zero value and zero delta, but nonzero gamma. So, to achieve the desired gamma using the combination of forwards in part b, requires only solving one equation in one unknown: $desired = $original balance sheet + Ncombination $combination . First, we need to calculate combination based on the net cash flows of the combination: $combination = 22 P V2 (6,934.25) 42 P V4 (2 8,285.68) + 10,000 10,000 + 62 P V6 (10,000) = 4.6971 . 10,000
balance sheet 2 P V (6,934.25) 6 P V (10,000) + = 234.85 100 100 2 2 2 P V (6,934.25) 6 P V (10,000) $ = + = 18.79 . 10,000 10,000 Since $desired = 35.62, $original = 106.47, $combination = 4.70, 35.62 = 106.47 + (4.70)Ncombination Ncombination = 30.25 Part b. Find NA and NB such that: 234.85 NA + 117.43 NB = 0 . There are an infinite number of solutions to this problem. One possible solution is: NA = 1 and NB = 2 (Note NB < 0 implies you need to short Forward Contract B.) Thus, you need to short 30.25 units of forward contract A, and you need to go long 2 30.25 units (or 60.50) of forward contract B. 2. Part a. You need to solve the following 4 equations and 4 unknowns: 775.64 26.24 23.27 20.64 NA NA NA NA + 254.40 NC + 5.42 NC + 4.94 NC + 4.51 NC + 393.79 ND + 58.74 ND + 39.38 ND + 26.40 ND + 633.06 NF + 38.92 NF + 29.98 NF + 23.34 NF = 100,000 = 500 = 1,000 = 1,500 Part c. If all wealth is invested in a 3year zero, then the number of units of a 3year zero (N3 ) that could be purchased is: N3 = 39,576.89/.775643 51,024.62 . = Such a holding of 3year zero coupon bond would result in a and of: 3 P V (1) = 1,187.31 $desired = N3 $3 51,024.62 = 100 32 P V (1) $desired = N3 $3 51,024.62 = 35.62 . = 10,000
Chapter 29: Solutions To Questions, Delta, Gamma, and Theta Page 1381 You could formulate the problem in other ways. However, you could only use 2 of the 3 of A, B, and C. Further, you could only use 2 of the 3 of D, E, and F . For example, B is redundant given A and C, for A and C can replicate the payoffs of B. Similarly, E is redundant given D and F , for D and F can replicate the payoffs of E. In my algebraic solution, I used A, C, D, and F . Of course, one could solve the problem using (say) B, C, D and E. Part b. Clearly, $ would seem to be negative, since $ increases as interest rates increase. (Recall $ is minus the slope of delta profile.)
Chapter 29: Solutions To Questions, Delta, Gamma, and Theta Page 1382 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Further, approx. minus slope 500  1,000 = 125 $ = = 48 of delta profile or  1,000  1,500 = 125 $ . = 8  12 Solutions to Characterizing Uncertainty about Bond Returns
1. Part a. Let F equal the forward price. This price should be set so that the present value of all the cash flows is zero. That is, 0 = F e2.083194 + 20e3.084688 + 20e4.086111 + 120e5.087465 . Solving the above equation for F : F = 126.58 . (Note the sign applied to the cash flows are from the perspective of the long, who pays F and receives the cash flows from the underlying security.) Part b. 1 ~ 1(126.58e1.051632 )  2(20e2.053194 )  3(20e3.054688 )  4(120e4.056111 ) $ = 100 =  3.5017 . Part c. 1 ~ 12 (126.58e1.051632 )  22 (20e2.053194 )  32 (20e3.054688 )  42 (120e4.056111 ) $ = 1002 =  0.1638 . Part d. V =126.58e1.051632  20e2.053194 )  20e3.054688 )  120e4.056111 ) =  10.62 . Part c. The numerical solution to the four equations and four unknowns in part a is: NA = 981.206 NC = 239.836 ND = 374.79 and NF = 1,180.96 . 1 V (10.62)  (3.5017)(6.6632  5.1632) + (.1638)(6.6632  5.1632)2 2  5.55 Chapter 29: Solutions To Questions, Delta, Gamma, and Theta Page 1383 Chapter 29: Solutions To Questions, Characterizing Uncertainty about Bond Returns Page 1384 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Solutions to Vasicek 1: Properties of the ShortTerm Rate
1. Clearly, the forecasts of the shortterm rate converge to 10.00%. So = 10%. Similarly, the standard deviation of the forecast error has converged to 6.00%, so = 6.00%. There are many ways to back out . We know (Forecast made at time s of t r) = + (s r  )ts .12 = .10 + (.15  .10)1 .12  .10 .02 = = = .40 . .15  .10 .05 Part b. 1. Part a. Solutions to Vasicek 2: The Term Structure 1N1 + .9814 N2 = .9441 1N1 + .9851 N2 = .9542 .9832 N1 + .9663 N2 = .9320 N1 = 1.7349 N2 = 2.7297 Expected Return on 9month bond Return from Buying 3month Bill
1/ 52 = (.5834) 1.41% . = = .9698 1 .9592  .9769  1 (1  .5834) .9592 2. Part a. .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... 5.47% .52 STEP = 2 ln h = .025 2 ln(.40) = 0.0046932 qV = 1/2 + (.06  .05) 1/52  ln(.40) .025 8 .52 = .5187728 = 1  1 = 1.32% . .9870 Since the expected return on 9month bill exceeds that of 3month bill, term premiums are positive. Part c.
c 0 d.5 5% .48 4.53% = .9851[(.65 .9832) + (.35 .9870)  (risk adjustment)] = .9694862 .25 [.9870  .9832] = .00038 Part b. values: Convert the continuously compounded yields to annually compounded e
.0547 Note: (risk adjustment) = .4/2  1 5.62% = e .0453  1 4.63% . = Part d. 1N1 + .9698 N3 = .9851 1N1 + .9769 N3 = .9889 .9870 N1 + .9592 N3 = .9734 N1 = 0.4661 N3 = 0.5352 Expected Annualized Rate on 1week zero with annual compounding: (.52 5.62%) + (.48 4.63) = 5.14% . Chapter 29: Solutions To Questions, Vasicek 1: Properties of the ShortTerm Rate Page 1385 Chapter 29: Solutions To Questions, Vasicek 2: The Term Structure Page 1386 Adventures in Debentures Adventures in Debentures Part e. 0 r 0,.25 rdown .25,.5 .25 = 4 ln(1/.9851) 6.00% = = 4 ln(1/.9870) 5.23% = Current cost: e0.098801 N1 + e.099592.5 N.5 + e.0980431.5 N1.5 = .823135 . Since perfect substitutes sell for the same price, the price of 2year paper in period 0.0 should be 0.823135. Such a price implies a yield of: .5 ln(.823135) = 0 r0,2 = 9.7317% . .25,.5 STEP = .25 rup  0 r0,.25 = 6.00  5.23% = 0.77% . Note: the calculation of the STEP will contain rounding error since the prices are only carried to 4 decimal positions. down, down .5 r .5,.75 Thus, E .75 r .75,1.0
E .75 d1.0 = 4 ln(1/.9889) = 4.46% . = 4.46%  .77% = 3.69% =
1 e.0369 /4 .9908 . = 3. Part a. STEP = 2 ln h = .025 2 ln(.70) .25 1.0558% = (Note: It is OK to infer the STEP from the tree by + .25 r .25,.5  0 r 0,.25 = 7.5558%  6.5000% = 1.0558%.) Part f. By inspecting the tree at t = .25, we see that
F .75 d1.50 = .9592 F .75 d1.75 A .5 r .5,.75 = 5.4442%  1.0558% = 4.3884% = .9447 . Part b. 1 (  .07558) h  ln + 2 8 1 (.08  .07558) .25  ln(.70) = + 2 .025 8 51.87% = Part g. 1N1 + .9795 N2 = .75 dG 1.50 1N1 + .9832 N2 = .9491 .9814 N1 + .9627 N2 = .9256 Thus,
G .75 d1.50 qV = N1 = 1.6491 N2 = 2.6426
B .25 d.25,.75 = .9393. = e.075558.25 [.5187 e.086115.25 + .4813 e.065.25  ( .2 2.25 )(e.065.25  e.086115.25 )] .96257999 = 2. As of time period 0.0, we can create a synthetic for the 2year zero coupon bond. The synthetic will contain the 6month zero, the 12month zero, and the 18month zero. To determine the number of units to hold in the synthetic: up: no change: down: e.5.110763 N1 + 1 N.5 + e.109527 N1.5 = e1.5.108349 e.5.099592 N1 + 1 N.5 + e.098801 N1.5 = e1.5.098043 e.5.088421 N1 + 1 N.5 + e.088074 N1.5 = e1.5.087737 . B .25 d.25,.75 = 1 e.5.25 r.25,.75 B B .25 r .25,.75 7.6276% = Part c. up: down: 1 N.25 + e.25.075558 N.5 = e.5.076276 1 N.25 + e.25.054442 N.5 = e.5.056104 N.25 = 0.8809462 N.5 1.8786806 = cost: e.25.065 N.25 + e.5.066190 N.5 = .9507767
C 0 d0,.75 = 1 e.750 r0,.75
C Solving 3 equations in 3 unknowns: N.5 = 0.634073 N1.0 = 2.22575 N1.5 = 2.59051 .
Page 1387 C 0 r 0,.75 6.7301% = Chapter 29: Solutions To Questions, Vasicek 2: The Term Structure Chapter 29: Solutions To Questions, Vasicek 2: The Term Structure Page 1388 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Solutions to Vasicek 3: More Term Structure
1. We need 3 results about Vasicek Model (for t s): As of s, Expected Value of t rt+h = + (s rs+h  )ts As of s, Std. Dev. of t rt+h = (for small h) 1  2(ts) (for small h) 2 2 v  yield on long term zero = rL = +  ln  ln 1  2 1 4 (1) (2) (3) (4) (5) 2. Part a. False. The local expectations hypothesis only assumes that the expected return for a short holding period is constant across bonds of all maturities. It does not imply term structure shifts are uniform. Part b. True. Even though returns on long term bonds are more volatile than the returns on short term bonds, the yields on long term bonds are less volatile. This apparent paradox is resolved by remembering that any change in the yield results in a price change reflecting the compounded value of the yield change. The analytical equation which describes the yield on any zero coupon bond using the Vasicek model could be used to verify this claim. Part c. False. The shape of the term structure reflects both term premiums as well as the expected direction of future short term interest rates. Even when term premiums are zero, short term interest rates may be expected to either increase or decrease in the future  resulting in nonflat term structure. Part d. False. Again, the shape of the term structure reflects both term premiums as well as the expected direction of future short term interest rates. When the term structure is upward sloping, interest rates may be expected to decline in the future if term premiums are large enough (i.e., large V ). Part e. False. Again, the shape of the term structure reflects both term premiums as well as the expected direction of future short term interest rates. When the term structure is upward sloping, local expectations hypothesis could be true as long as interest rates are expected to increase in the future. If the local expectations hypothesis is true, then the expected returns from short term bonds and long term bonds are equal. Part f. Uncertain. The answer is uncertain because it is not clear what is being assumed about the pricing of longterm bonds relative to shortterm bonds. For example, if local expectations is true, then the expected return from short bonds should be the same as expected return from longer bonds. In this case, the firm would never anticipate additional profit due to the shape of the term structure. If the expected return on longer bonds is (say) greater than the expected return from shorter bonds, then the institution may anticipate a profit from lending longterm and borrowing shortterm. But the shape of the term structure does not necessarily imply the existence of a term premium. The shape reflects both a term premium (if Bond Managers Forecast .0217945 = of uncertainty .0293215 = Bond Managers Forecast .087 = + (.09  ) of interest rates .0843 = + (.09  )2 2 2 v  . Finally, Long Term yield .1451941 = +  ln  ln Now, we need to solve for , , , V based on the first 5 equations. (Even if you don't find the numeric values for , , , V , recognizing that you need to examine equations (1)(5) will give you most of the points.) Equation (3) can be solved to find: (.09) = (.087)/. Substitute this result into equation (4) to find .0843 = + (.087  ). Using this last equation and equation (3) indicates: .087 = (  ) + .09 .0843 = (  ) + .087 . Subtract the second equation from first which implies .0027 = .003 = .9. Given = .9, use equation (3) to find = 6%. Use equation (1) or (2) and = .9, to find = 5%. Use equation (5) and = 6%, = .9, and = 5%, to determine v = .5.
Chapter 29: Solutions To Questions, Vasicek 3: More Term Structure Page 1389 Chapter 29: Solutions To Questions, Vasicek 3: More Term Structure Page 1390 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons it exists) as well as the market's expectations about future interest rates. Part g. True. As the speed of mean reversion decreases, this implies that is getting closer to one. As gets closer to one, any change in the shortterm yield is regarded as more permanent. Thus, the longterm yield is affected more since the change is permanent. 1. Part a. 3. Part a. Solutions to Vasicek 4: The Greeks STEP = qV,t STEP = .03 2 ln(.45) 1 = .0379 1 (.16  .14) 1  ln(.45) qV = + = .71 2 .03 8
. ... ... .. .. ... ... .. .. 2 ln h = .05 2 ln(.30) .25 = .0387939 h  ln 1 (  t r) = + 2 8 = .8879389  3.8793891 t r
. . ... ... ... ... ... ... .1779
.. .. .. ... ... ... .. .. 22.75878% .14 ... ... .. .. ... ... ... ... ... ... . . ....... ... ... ... ... ... ... ... ... .. . .29 .............. ... ... ... ... . .71.............. .1555................. 18.87939% .3060.................. .1021 15% .6940 .. ... ... ... .. .. ... ... ... ... . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .8445 .4565 Part b. 0 d2 = e.14 [.71 e.1779 + .29 e.1021 ] .7443 = 11.12061% .5435 . .. ... ... ... ... ... .. .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . ... ... ... ... ... ... .. ... ... ... .. ... ... ... . .. ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... 15% 7.24122% 4. up: down: 699.55 N3 + (979.92 + 65.37)NB = 1,000.00 750.54 N3 + (1,026.01 + 65.37)NB = 1,000.00
1,000 = 2 ln( 941.62 ) 12.03% = N3 = 2.1888128 NB = 2.42151361 Part b. Risk Adjustment Factor =
d u v h et rt+h h  et rt+h h 2 .4 .25 .1112061.25  e.1887939.25 e = 2 = .0018683 cost: 676.03 N3 + 999.92 NB 941.62 = 0 r 0,.5 NOTE: Knowing the probability of an up move is 50% plays no role in the solution.
0 d.5 = e.15.25 .3060305 e.1887939.25 + .693965 e.1112061.25  .0018683 = .9294782 Chapter 29: Solutions To Questions, Vasicek 3: More Term Structure Page 1391 Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1392 Adventures in Debentures Adventures in Debentures Since
t dt+2h Summary of Evolution of Discount Function: = et rt+2h 2h 14.63% . 0 r .5 =
.. .. .. .. ... ... ... ... ... ... .. ... ... ... . ... ... ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .25 d.5 .25 d.75 Risk Adjustment Factor = .1 e.15.25  e.2275878.25 = .0018503
u .1887939.25 .1555339 e.2275878.25 + .8444661 e.15.25 .25 d.75 = e 0 d.25 0 d.5 0 d.75  .0018503 = .9142791 u .25 r .75 = .9632 = .9295 = .8982 = .9539 = .9143 17.92% = .25 d.5 .25 d.75 = .9726 = .9449 .. ... ... ... .5 .. .. ... ... ... ... ... ... .. ... ... ... ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . ... ... .5 .. ... ... ... .. ... ... ... .. ... ... ... . .. ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... d.75 = .9447 d.75 = .9632 .5 d.75 = .9821 Risk Adjustment Factor = .1 e .0724122.25 e .15.25 = .0018865 Part c. Initial Term Structure: "Up" Term Structure: 0 r .25 .25 r .5 = 15% = 18.88% 0 r .5 .25 r .75 = 14.63% = 17.92% d .1112061.25 .4565272 e.15.25 + .5434728 e.0724122.25 .25 d.75 = e  .0018865 = .9449219 d .25 r .75 = 11.33% . To determine 0 d.75 create a synthetic 9 month bond: up: down: current value of synthetic 1N.25 + e.1887939.25 N.5 = .9142791 1N.25 + e
.1112061.25 "Down" Term Structure: N.25 = .6502 N.5 = 1.6401 0 r .75 = 14.32% .25 r .5 = 11.12% .25 r .75 = 11.33% N.5 = .9449219 If Term Structure Makes Uniform Shifts, then: "Up":
.25 r .75 = e.15.25 N.25 + .9294782 N.5 = .8981596 = 14.63 + (18.88  15) = 18.51% = 17.92% "Down": .25 r .75 = 14.63 + (11.12  15) = 10.75% = 11.33% Thus, Vasicek model does not generate uniform shifts. Summary of Evolution of Term Structure: .. ... .. .. .5 ... ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . .. .. ... ... .5 ... ... ... ... .. ... ... ... . .. ... ... ... ... ... ... .. . ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... r.75 = 22.76% .. .. .. ... = 15% ... ... .. .. ... ... .. .. ... ... ... . 0 r .5 = 14.63% ............. ... ... ... ... ... ... ... ... 0 r .75 = 14.32% .. .. 0 r .25 . ... ... ... .. ... ... r .. .25 .5 .. ... ... .25 r .75 = 18.88% = 17.92% Part d. This question provides an example of the type of question that is addressed in horizon analysis.1 In period 0, r.75 = 15% investor buys 40% of 1000 in 9 month bonds N.75 = 400 /0 d.75 = 445.35515
1 ... ... ... ... ... ... . .25 r .5 .25 r .75 = 11.12% = 11.33% .5 r .75 = 7.24%
Page 1393 Horizon analysis is just a simulation of future possible returns based on different strategies for constructing a bond portfolio. The simulation is across various possible scenarios for alternative future term structures. Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1394 Adventures in Debentures Adventures in Debentures investor buys 35% of 1000 in 6 month bonds N.5 = 350 /0 d.5 = 376.55536 investor buys 25% of 1000 in 3 month bonds N.25 = 250 /0 d.25 = 259.553 Part f. Current Value = 25 e.15.25 + 125 e.1463.5 = (25 .9632) + (125 .9295) = 24.08 + 116.18 140.26 =
Payoff of Portfolio in Up State = (445.35515 .9142791) + (376.55536 .9538981) + (259.553 1) = 1025.9273 $ = .25 P V (25) + .5 P V (125) = (.25 24.08) + (.5 116.18) = 64.11 Return = 2.59273% Annualized Return = (1.0259273)4  1 10.78% = Payoff of Portfolio in Down State = (445.35515 .9449219) + (376.55536 .9725814) + (259.553 1) = 1046.6096 V = 25 V.25 + 125 V.5 Return = 4.66096% Annualized Return = (1.0466096)4  1 19.99% = where V T = Vasicek Delta for a T period zero coupon bond e.1887939.25  e.1112061.25 = .002408024 18.87939  11.12061 V = (25 0) + (125 .002408024) .3010 = V.5 =  V.25 = 0 Expected Annualized Return = (10.78 .3060) + (19.99 .6940) = 17.17% Std. Dev = (10.78  17.17)2 (.3060) + (19.99  17.17)2 (.6940) = 4.24% Thus, V < $ as expected. Alternative way to calculate V for bond Part e. Expected Value of .25 r .75 = (17.92 .3060) + (11.33 .6940) = 13.35% B + = 25 + 125 e.1887939.25 = 144.23726 B  = 25 + 125 e.1112061.25 = 146.57267 B+  B 146.57267  144.23726 V =  + = = .3010 r  r 18.87939  11.12061 V for this bond in 3 months is zero because the $125 final payment is received for certain at the next time period. The forward rate is: 0 r .25,.75 = 1 (.75 0 r0,.75  .25 0 r0,.25 ) = (1.5 14.32%)  (.5 15%) = 13.98% . .5 Clearly, the forward rate is not equal to the expected spot rate!
Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1395 2.
Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1396 Adventures in Debentures Adventures in Debentures Part a. V 3 = (+3  3 ) V V r+  r +3 = V = (2 d++  2 d+ ) 3 3 r++  r+ (e.123444  e.07 ) 12.3444  7.0 Part c. VNE = 59,159.5275 = .81680945 N3 + .76713328 N4 + .72181129 N5  (.81680945 50,000) V,NE = 632.927793 = .01549756 N3 + .01947203 N4 + .02190902 N5  (.01549756 50,000) V,NE = 0 = .00009326 N3 + .00027122 N4 + .00044913 N5  (.00009326 50,000) [The question only asks you to specify the 3 equations in 3 unknowns (N3 , N4 , and N5 ) to be solved. Just for your information, the solution to the above system of equations is N3 = 347,679.4; N4 = 420,751.8 and N5 = 192,273.1.] Part d. If you invest in a 2 year zero coupon bond, the $59,159.5275 will grow to (59,159.5275/0 d2 ) = $67,873.95. (You can verify that the solution for N3 , N4 and N5 given in part c will also provide the same net equity in year 2.) 3. Part a. The implied forward rate is: 10,000 e0 r2,3 = 10,760.03 0 r2,3 = 7.3253% = .00907916 3 = V = (2 d+  2 d ) 3 3 r+  r (e
.07 e 7  1.6556 .016556 ) = .00957759 V 3 = (0.00907916  .00957759) 9.6722  4.3278 = 0.00009326 Part b. If you invested all of your equity in a 2 year zero, then the value of the equity in 2 years is known with certainty. Such an investment would have the following characteristics: Current Value of Equity = $100,000  (50,000 0 d3 ) = 100,000  (50,000 .81680945) = 59,159.5275 To purchase $59,159.5275 worth of 2 year zeros, requires one to purchase (59,159.5275/.87160879) units of the 2 year zero. The V for such an investment is: 59,159.5275 .00932505 = 632.927793 . V equity = .87160879 The V of a 2 year zero is 0, so V equity must equal 0 as well.
Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1397 If you wanted to confirm the forward agreement has zero value as of year 0: (.84449 10,000)  (.78484 10,760.03) 0 . = Part b. Value of Forward Borrowing = 10,000  (.84294 10,760.03) = 929.94 . in Year 2 (upup case) For Part c, it is useful to also calculate: Value of Forward Borrowing = 10,000  (.91393 10,760.03) = 166.09 . in Year 2 (updown case) Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1398 Adventures in Debentures Adventures in Debentures Part c. Payoff to the forward position:
. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . .. ... ... ... ... ... ... ... ... .. .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . Part a. 929.94
. ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... . . .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . 166.09 To find the synthetic version of the payoffs for this forward contract: 1.00000 N1 + .73986 N3 = 929.94 1.00000 N1 + .84449 N3 = 166.09 N1 = 6331.28 N3 = 7300.49 .
1 d2 0V 1 d3 C = .8607 = .7432 = 8,723.59 The value of this synthetic (as well as the forward contract) is: .87772 N1 + .73036 N3 = 225.11 . .. .. .. .. .. .. .. .. .. .. . . .. .. .. .. .. .. .. .. .. .. 1 .. .. .. .. . . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . ... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . = .8509 = .7272 bond's coup = 400 V A = 10,037.51
1 d3 1 d2 2 d3 = .8412 cash from bond = 11,663 + 428 = 12,091 Face Coupon 2 d3 = .8607 cash from bond = 11,235 + 428 = 11,663  Note: these do
. ... ... .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... . .. .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . not match. = .8607  cash from bond = 10,815 + 412 = 11,227 2 d3 NOTE: I did not give you the value for 1 du to force you to solve this question by 3 finding a synthetic. If you knew 1 du , an alternative solution is: 3 (.87772 10,000)  (10,760.03 1 du ) . 3 You can verify that you get the same value (within rounding error) for the forward contract if I had told you 1 du = 0.79481. 3 Part d. Note: up, up 2 r 2,3 = ln(1.00000/.84294) = 17.09% V Forward up, down 2 r 2,3 = ln(1.00000/.91393)
tV = .8706 = .7595 bond's coup = 400 B = 9,564.63 1V
1 d3 1 d2 2 d3 = .8807 cash from bond = 10,403 + 412 = 10,815 Z market price (full or flat) of inflationindexed bond in scenario Z at year t. The market price at date t excludes coupon payment made at date t. = 9.00% 929.94  (166.09) = 94.52 . = 17.09  9 NOTE: This inflationindexed bond provides an example of a path dependent security. Path dependent securities have the property that some of the inner nodes in the relevant tree diagram fail to touch  that is, going up at year 1 and down at year 2 is not the same as going down at year 1 and up at year 2. up: down: 4.
Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1399 1 N1 + .8412 N2 = 12,091 1 N1 + .8607 N2 = 11,663 N1 30,554.26 = N2 21,948.72 = cost: .8509 N1 + .7272 N2 = 1 V A 10,037.51 =
Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1400 Adventures in Debentures Adventures in Debentures up: down: 1 N1 + .8607 N2 = 11,227 1 N1 + .8807 N2 = 10,815 N1 28,957.42 = N2 20,600.00 = Part c. The following tree measures the additional cash flow if we move the coupon yield up by 100 basis points.
.. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. .. ... ... .......... . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 107 cost: .8706 N1 + .7595 N2 = 1 V B 9,564.63 = up: down: 1 N1 + .8509 N2 = 1 V A + 400 = 10,437.51 1 N1 + .8706 N2 = 1 V B + 400 = 9,964.63 N1 30,862.57 = N2 24,004.06 = cost: .8607 N1 + .7432 N2 = 0 V C 8,723.59 = (10,037.51 + 400)  (9,564.63 + 400) = 206.61 16.1461  13.8573 . .. .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... . .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . 100 107 103 100 Part b. V =  .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... . .. ... .. .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 103 where 16.1461%  ln(.8509) and 13.8573%  ln(.8706). = = V = V
0V C Valueadded as of up scenario at t = 1 = .8509 107 = 91.0463 Valueadded as of down scenario at t = 1 = .8706 103 = 89.6718 For t = 0: up: 1 N1 + .8509 N2 = 100 + 91.0463 = 191.0463 down: 1 N1 + .8706 N2 = 100 + 89.6718 = 189.6718 N1 250.4149 = N2 69.7716 = = 206.61 = 2.3684% 8,723.59 In understanding the numeric value of V , there are two issues. First, V is negative because of the lag built into the indexation for the bond. When rates go down in period 1, this is normally good news for bondholders. However, in this example when rates drop in year 1, inflation also drops between years 0 and 1. This drop in the inflation has a significant impact on the nominal cash flows of this indexed bond in year 2. This indexation effect offsets the drop in nominal rates leading to a decrease in the price for the indexed bond, not the usual increase! Second, the risk of the indexed bond (as measured by the absolute value of V ) is greater than the risk of a comparable bond that is not indexed. Again, this should not be surprising upon reflection. V measures the price sensitivity of a bond to changes in interest rates in nominal terms, not real terms. The fact that the inflationindexed bond is more volatile in nominal terms does not necessarily mean it is more risky in real terms. To illustrate this idea, we could calculate V using the price sensitivity in real, not nominal, terms. Assume the economy's price index is 100 at t = 0, and 107 in the up scenario at t = 1, and 103 in the down scenario at t = 1. In this case, the "omega" for the inflation indexed bond is:  9,564.63+400 8,723.59 103 = 0.402%. "V " =  16.1461  13.8573 100 In contrast, a similar methodology for the second bond, which is not indexed to inflation, gives its "omega" to be 2.89%. Now, the magnitude of the inflationindexed bond is closer to zero.
10,037.51+400 107 valueadded .8607 N1 + .7432 N2 163.6779 = as of t = 0 When the coupon yield is 4%, part a demonstrates the value of the bond is 8,723.59. For bond to sell for par, we need the price to increase by 10,0008,723.59 = 1,276.41. For every 100 basis point increase in the coupon yield, we expect the value to increase by 163.6779. Thus, we need to increase the coupon yield by 1,276.41 = 7.7983 163.6779 or set it equal to 11.7983%. You can verify that with a coupon yield of 11.7983%, the value of the inflationindexed bond is 10,000when you use the methodology of part a. Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1401 Chapter 29: Solutions To Questions, Vasicek 4: The Greeks Page 1402 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Solutions to Introduction to Bond Options
1. Part a. This is a straightforward application of putcall parity. We know ct = X Bt + pt  (1+r)T t , where ct is the current price of the call option, Bt is the current price of the underlying bond, pt is the current price of put, X is the exercise price for put and call, r is the interest rate, and (T  t) is the amount of time before expiration of the put and call. For this problem, putcall parity requires that the value of the call be 40 ct = 40 + 2  1 = 2.65, (1.05) 3 but the call is selling for $3 > 2.65. The call is over valued. To create an arbitrage profit, we need to sell the call and buy the portfolio which replicates the call. The replicating portfolio consists of the underlying bond, 1 put, and borrowing the present value of exercise price. To show the arbitrage profit, we construct the following table: Cash Flow Today Write 1 call Buy 1 put Buy 1 underlying bond Borrow 39.35@5% NET +3 2 40 +39.35 +0.35 Cash Flow at Expiration of Options if BT 40 if BT < 40 0 40  BT BT 40 0 (BT  40) 0 BT 40 0 interest rate by r, and the number of periods before the option matures by (T  t). That is, X . ct = Bt + pt  (1 + r)T t However, the problem does not give you enough information to apply directly the above equation, for you are not given a put with the same exercise price as the call. The putcall parity relation assumes that the exercise prices of the European call and the European put are the same and equal X. Fortunately, you do know that the value of a put must be nonnegative since a put is a limited liability security. Thus, putcall parity does imply: X . ct Bt  (1 + r)T t The right hand side of the above equation is equal to 60  [40/(1.11111)] = 24. If the price of the call is equal to 22, then the above inequality is violated. To arbitrage this situation, we wish to buy the call because it seems to be under priced according to putcall parity. Thus, consider the payoffs from a portfolio which consists of buying 1 call, lending $36 at the interest rate of 11.111%, and shorting the underlying bond: Cash Flow Today Buy Call Short Underlying Bond Lend NET 22 +60 36 +2 Cash Flow at Expiration of Option if 40 < BT if BT < 40 0 BT +40 40  BT > 0 +(BT  40) BT +40 0 The minimum profit (0.35) occurs if you buy only one portfolio. However, profit remains available as long as the prices do not adjust to eliminate the arbitrage. Thus, the minimum profit occurs if you buy only 1 portfolio like this, and it turns out that the price of the underlying bond at the expiration (denoted by BT in the above table) of the option is greater than $40. Part b. We know from putcall parity that the current price of a European call option (ct ) must equal the sum of the current price of the underlying bond (Bt ) and the current price of the European put option (pt ), less the discounted value of the exercise price. Throughout we will refer to the current exercise price by X, the
Chapter 29: Solutions To Questions, Introduction to Bond Options Page 1403 Chapter 29: Solutions To Questions, Introduction to Bond Options Page 1404 Adventures in Debentures Adventures in Debentures 2. Using the same notation and using putcall parity, we know that the current price of the underlying bond must be $100. This follows from noting that the value of a put with a zero exercise price must be zero and noting the discounted value of the exercise price is zero since the exercise price is zero. Under these circumstances, putcall parity implies ct = Bt , or 100 = Bt . Now, let's look at the implications of putcall parity using Bt = 100: ct (X = 10) = 100 + pt (X = 10)  ct (X = 20) = 100 + pt (X = 20)  10 1.11111 20 1.11111 = 91 + pt (X = 10) = 82 + pt (X = 20) 3. To solve this problem we use putcall parity. We know: 88 100  (1 + r2 )2 1 + r1 100 91 c(X = 91) = p(X = 91) +  . (1 + r2 )2 1 + r1 c(X = 88) = p(X = 88) + We have 2 equations in 2 unknowns. Substitute in the prices for puts and calls and subtract the second equation from the first: 88 100  (1 + r2 )2 1 + r1 100 91 0.0364 = 1.6045 +  . (1 + r2 )2 1 + r1 1.4001 = 0.2409 + This implies r1 = 9.9989%. To solve for r2 , note: 1.4001 = 0.2409 + This implies r2 = 11.0015%. 88 100  . (1 + r2 )2 1.09989 The problem states that c(X = 10) equals $93, which implies by the above equation that p(X = 10) equals $2. The problem also states that c(X = 20) equals $83, which implies by the above equation that p(X = 20) equals $1. But we know that a put with a higher exercise price must sell for a higher price than one with a lower exercise price. Thus, the calls cannot be properly priced in the market place. To arbitrage this situation, we wish to buy the call with an exercise price of $20 because it seems to be under priced according to putcall parity and sell the call with an exercise price of $10 because it seems to be overpriced using the putcall parity argument. Thus, consider the payoffs from a portfolio which consists of buying 1 call with exercise price of $20, selling 1 call with an exercise price of $10, and lending $9 at the interest rate of 11.111%. Cash Flow Today Buy c(X = 20) Write c(X = 10) Lend $9 NET 83 +93 9 +1 Cash Flow at Expiration of Option if 20 < BT if BT < 10 if 10 BT 20 4. Part a. Harpo should exercise. If Groucho found the benefit of the coupon exceeds the loss of insurance and the time value of money, then Harpo must reach the same conclusion because Harpo has less insurance (i.e., a larger deductible) and the time value of money on 85, not 90. Part b. Chico may or may not exercise. More information is needed. 0 0 +10 10 0 (BT  10) +10 20  BT 0 +(BT  20) (BT  10) +10 0 Part d. Karl should exercise. Karl's position is a combination of Harpo's and Zeppo's. If it is optimal for Harpo, Groucho, and Zeppo to exercise early, it must be optimal for Karl to exercise early. Part c. Zeppo should exercise. Zeppo's position is similar to Groucho's except shorter maturity. Thus, the loss of insurance and the time value of money is smaller for Zeppo than Groucho. Thus, the minimum profit occurs if you buy only 1 portfolio like this, and it turns out that the price of the underlying bond at the expiration (denoted by BT in the above table) of the option is greater than $20. Chapter 29: Solutions To Questions, Introduction to Bond Options Page 1405 Chapter 29: Solutions To Questions, Introduction to Bond Options Page 1406 Adventures in Debentures Adventures in Debentures 5. Part a. Part c. To duplicate the payoff of a variable price forward note that Possible Market Prices 97.0 98.9 99.0 99.1 99.5 100.0 101.0 Seller's Action Value of Buyer's Position i. writing a put with an exercise price of 100 and a  Seller sells at 99  Buyer's Position Worth 2.0       " " " " " " " " " " " " " " " " " " " " " "      " " " " " " " " " " " " " " " " " " 0.1 0 +0.1 ii. buy a zero coupon bond +0.5 +1.0 +1.0 with face value of $1 that matures in 60 days implies:
. . . Value ..... . . . . . . of . . . . . . . Zero . . . . . . . on . . . .................................................. ................................................... . . . Maturity . . .. . . ... . . . . . . . ... $1 . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . .. . . . . . . . . . Value ..... . . . . . . of . . . . . . Put . . . . . . . . on . . . . . . Maturity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . .. maturity of 60 days implies: ....... ...... ............................................................... TNote .... Price .... .... .... .... .... ... 100 " 100  . . ...... ...... . TNote Price Part b. This payoff diagram is like the payoff from writing a put option except for the $1 value when TNote price . Value ..... . . . . of Buyer's Position . . . iii. do (i) and (ii) together: 1 0 is greater than or equal to $100. (Usually the maximum payoff from writing a put is zero.)
1 .................................. ................................... .. ... . . ... ... ... . . .. . ... . . ... . ... . .. . ... . . ... . $1 ... . .. . ... . . ... ... .. . .. .. ... . .. ... ... . .. . ... ...... ...... . ... ... .. .. ... ... 99 98 100 Price ... ... ... ... .. of ... ... ... .. ... TNote ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... . . ... ... . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . .. ................................................ .... .... . . ...... ...... . .... 100 TNote .... 99 .... Price .... .... .... .... .... Conclusion: (i) and (ii) are a synthetic variable price forward. Therefore, the current value of a variable price forward should be .96  .35 = .61. Chapter 29: Solutions To Questions, Introduction to Bond Options Page 1407 Chapter 29: Solutions To Questions, Introduction to Bond Options Page 1408 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures 2. Part a. The long position of the forward contract has the following cash flows: Solutions to European Bond Options
1. Part a. .9704455  .965 = 0.0054455 0 1 2 pay 920,378.18 3 receive 1,000,000 where 1,000,000 = 920,378.18 . 1.0865099 Thus the values of the forward contract in year 0 and year 1 are: Part b. 1N1 + .9631944 N2 = .0005219 1N1 + .9680225 N2 = .0043578 0 .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... . .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . .  920,378.18 1,000,000 = 19,511.06 + 1.1238 (1.1184)2 Solving the above 2 equations: N1 = .7647310 N2 = .7944974 . Year 0  920,378.18 1,000,000 = 15,219.48 + 1.0653 (1.0665)2 Year 1 The cost of this replicating portfolio as of period 0 is: .9656054 N1 + .9340284 N2 = .00365228 . Clearly, you would not exercise the option to acquire the long position of the forward contract if interest rates go up in year 1. Thus, the values of the call option on the long position of the forward contract are .. .. ... ... 0 ... ... ... ... ?
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . Part c. V =  .0005219.0043578 = .001918 15.0013.00 Note: To find the 13% and 15% in the above denominator we solve: 4 ln[1/.9631944] 15% and 4 ln[1/.9680225] 13% . = = 15,219.48 Year 1 Year 0 Part d. .00365228 (same value as in part b). Based on our discussion of putcall parity, we know early exercise of an American call option on a zero coupon bond is not optimal. Thus, values of the European call option and the American call option are the same. Part e. Use putcall parity: p = c + P V (X)  B = .00365228 + (.965 .9340284)  .9049280 = .0000616 . To solve for ?, find a portfolio of 1 and 2 year bonds that replicates the payoffs of the call option: 1 N2 = 0 1N1 + 1.1238 1 N2 = 15, 219.48 . 1N1 + 1.0653 The solution to the last two equations is: N1 = 277,150.63 and N2 = 311,461.88 . Chapter 29: Solutions To Questions, European Bond Options Page 1409 Chapter 29: Solutions To Questions, European Bond Options Page 1410 Adventures in Debentures Adventures in Debentures The cost of the replicating portfolio is: 1 1 N2 = 7,805.55 . N1 + 1.0942 (1.0922)2 Thus, the cost of the call option in year 0 is 7,805.55. The value of this synthetic at B is: .9324 N1 + .8639 N2 = 0.0077 . Part b. Clearly, the option to exercise in year 0 is never used, for the forward contract is only worth zero in year 0 while the option (if not exercised) is worth 7,805.55. Thus, the value of the American call option is 7,805.55, which is the same as the European call option. NOTE: Asianstyle options are "path dependent." That is, to value this security you need to know both the scenario as well as how you arrived to a given scenario. You can see in the above tree that going up and then down is not the same as going down and then up (that is, the path you follow matters). We will encounter more path dependent securities later in the course. Part b. Note: up 1 r 1,2 = ln(1/.9078) = 9.67% down 1 r 1,2 = ln(1/.9576) = 4.33% 3. Part a. The following tree is relevant for this option:
. .. .... .... .... .... .... .... .... .... .... .... .... .... .... . . .... .... .... .... . .. . ............. .... .... .... .... .... .... .... .... .... .... .... .... .... .... .. .. . . ... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... . .... ... ............. .... .... .... .... .... .... .... .... .... .... .... .... .... .... .. .. 0  0.0219 = 0.0041 . V =  9.67  4.33 MAX{0; 1/3(.8839 + .8238 + .7968)  .86} = 0 B . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . 0 MAX{0; 1/3(.9324 + .8238 + .7968)  .86} = 0 MAX{0; 1/3(.9324 + .9058 + .7968)  .86} = .0183 A Part c. The value of a similar option (but not Asian style) would be greater. Your explanation could be based on either of the following ideas: First, a Europeanstyle option would have nonzero values for all scenarios in year 2. Further, the nonzero values would have larger values relative to its Asianstyle counterpart. These greater payoffs lead to greater current values at t = 0. Second, an Asianstyle option is based on an average which will have less volatility than the price in year 2. Options have greater value as the volatility of the underlying asset increases (holding everything else constant). Hence, the Europeanstyle option should have greater value than its Asianstyle counterpart. MAX{0; 1/3(.9835 + .9058 + .7968)  .86} = .0354 To find the value at A, we find a synthetic giving the same payoffs: 1N1 + .9324 N2 = .0183 1N1 + .9835 N2 = .0354 The value of this synthetic at A is .9576 N1 + .9058 N2 = 0.0219 . To find the value at B, 1N1 + .9078 N2 = 0.0000 1N1 + .9576 N2 = 0.0219 N1 = 0.3992 N2 = 0.4398 .
Page 1411 Chapter 29: Solutions To Questions, European Bond Options Page 1412 N1 = 0.2937 N2 = 0.3346 . Chapter 29: Solutions To Questions, European Bond Options Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Assets Liabilities 4,000 units of calls Market Value = 4,000 2.44 = 9,760 Solutions to American Bond Options
1. Value of coupon bond at (t = 0) is (20 .9091) + (120 .8116) = 115.57 Value of coupon bond at (t = 1, up) is 120 .8747 = 104.96 Value of coupon bond at (t = 1, down) is 120 .9108 = 109.30 1yr. zero = 0 Calculate r+ and r based on 1 d+ = .8747 and 1 d = .9108: 2 2 100 units of coupon bond Market Value = 100 115.57 = 11,557 10,000 units of 1 yr. zero Market Value = 10,000 .9091 = 9,091 Total Assets = 20,648 Net Equity = 20,648  9,760 = 10,888 VNE = 10,888 = 11,557 + .9091 N1 + .2469 Np  9,760 .9091 N1 + .2469 Np = 9,091 NE = 0 = (100 1.0697) + 0 N1  .1348 Np  4,000(.7023) .1348 Np = 2,702.23 Np 20,046.22 = .9091 N1 + (.2469 20,046.22) = 9,091 N1 15,444.30 = .8747 = er r+ 13.39% =
+ and .9108 = er r 9.34% =  Summary: Write 20,046.22 units of puts (i.e., liability) purchase 15,444.30 units of 1 year zero. 2yr. zero = (.8747  .9108) = .008914 13.39  9.34 2. Part a. B2 = 100 1.13 100 .4 + 1.06 .6 = 83.6382 1.10 call =  1.2676  4.112 = .7023 13.39  9.34 put =  .5458  0 = .1348 13.39  9.34 Part b.
+ B2 =  B2 = 100 1.14 100 1.1 100 .3 + 1.10 .7 = 79.6037 1.13 100 .65 + 1.04 .35 = 87.4951 1.06 Potential Prices of 2 period bond next period . . .... .... coupon bond = ..20.......1yr. + 120 2yr. zero .. .. .... .... . .... ..... 0 .... . .... = 120 .008914 = 1.0697
Chapter 29: Solutions To Questions, American Bond Options Page 1413 Chapter 29: Solutions To Questions, American Bond Options Page 1414 Adventures in Debentures Adventures in Debentures To find the current price of the three period bond, we follow a strategy of replicating the future cash flows of the three year bond. We use the one and two year bonds to create the synthetic three year bond. That is, find the solution: upstate: downstate: 100 N1 + 100 N1 +
100 1.13 100 1.06 N2 = 79.6037 N2 = 87.4951 N1 = 0.3989 N2 = 1.3503 3. In year 1, the down scenario, early exercise of the American put is not rational since .9058 is greater than the exercise price of 0.87. In year 1, the up scenario, early exercise of the American put is rational since .87  .8238 = .0462 > .0295. (.0295 is the value if you do not exercise, and .0462 is the value of exercising early.) To value the put in year 0, we need to solve 2 equations: 1 N1 + .9078 N2 = .0462 1 N1 + .9576 N2 = .0054 N1 = 0.7899 cost of replicating portfolio = 100 N1 + 83.6382 N2 = 76.6714 1.10 N2 = 0.8193 Part c. Payoffs to European put:
. .... ... .... .... .... .... .... .... .... .... .... .... .... .... ... ... .... .... .... .... .... .... .... .... .... ... . ............. .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ... ... cost of = .9324 N1 + .8639 N2 = 0.0288 . synthetic 90 
100 1.13 = 1.5044 In year 0, early exercise is not desirable since the exercise price .80  .7968 = .0032 < .0288. (You must note this to get full credit.) Thus, the value of the put in year 0 is 0.0288. 0 Today Next Year 4. Part a. There are two ways to approach this problem. The first approach would involve starting at time period t = 2 and working backwards. At each node you would find the synthetic by solving 3 equations and 3 unknowns. A second approach would rely on synthetics based on putcall parity. The second approach involves very few calculations, so we will follow the second approach here. Putcall parity claims: c = B + p  P V (x) = (1 .694068) + (1 .013146)  (.90 .784793) This suggests a synthetic alternative to the explicit call would consist of: 1 unit of the 3year zero coupon bond (i.e., the "underlying" bond); 1 unit of the put, and going short (or issuing) 0.90 units of the 2year zero coupon bond. (Note: at time period 0, the maturity of the explicit call is 2 years, so we need to create the leverage in the synthetic with a 2year zero.)
Chapter 29: Solutions To Questions, American Bond Options Page 1416 upstate: downstate: 100 N2 + 79.6037 N3 = 1.5044 1.13 100 N2 + 87.4951 N3 = 0 1.06 N3 = 0.61 N2 = 0.56 Current price of = 83.6382 N2 + 76.6714 N3 = 0.55 European put Part d. From part c: N2 = 0.56 units of 2 period bond N3 = 0.61 units of 3 period bond .
Chapter 29: Solutions To Questions, American Bond Options Page 1415 Adventures in Debentures Adventures in Debentures Part b. Clearly, the market value of the synthetic call (as of year 0) is: c = .694068 + .013146  (.90 .784793) = 0.0009003 guaranteed to be worth $100 on period 1.0. (Either the digital put or the digital call will expire in the money and pay $100.) Such a portfolio is a synthetic zero coupon bond with a face value of $100 and a maturity date of period 1.0. If perfect substitutes sell for the same price, then ct + pt = Bt . Since digital options have limited liability, ct 0 and pt 0. Thus, ct Bt and pt Bt . 5. Part a. V =  V = 92.1973  93.2605 = .67291139 10.79  9.21 V .67291139 = = 0.74% 90.3938 90.3938 94.3697  95.1066 + =  = .46639241 V 11.58  10.00  =  V V =  = 95.1066  95.8492 = .47000000 10.00  8.42 +   .46639241  .47000000 V V = = .00228329 10.79  9.21 10.79  9.21 Part d. From part c, we know that a portfolio containing 1 digital call and 1 digital put is equivalent to a portfolio which holds a zero coupon bond that matures on period 1.0 (with a face value of 100). Since these 2 portfolios are equivalent, the characteristics of each portfolio should be identical. Thus, ct + pt = Bt c + p = B p = c =
.... . . pt = Bt  ct = 90.3938  21.6557 = 68.7381 p = B  c = .6729  18.6265 = 17.9536 p 17.9536 = 26.12% = pt 68.7381 c 18.6265 = 86.01% = ct 21.6557 c = B  p = .0023 + 10.6656 = 10.6679 c = B  p = 9.4256  14.6122 = 5.1866 95.1066  90.3938 = 9.4256 2 .25 c + p = B c + p = B
. . . . . . . . . . . . . . . . . . . . . Part b. Value of Digital Call at Expiration . . . .. . . ..... . . . . . . . . . . . . . . . . . . . . . . Value of Digital Put at Expiration ...... ..... .. . .. .. ....... . .. 93 Value of Underlying Bond at Expiration of Option 93 Value of Underlying Bond at Expiration of Option Part c. By inspecting the payoff diagrams from part b, it should be clear that if you have a portfolio containing 1 digital call and 1 digital put, then the portfolio is
Chapter 29: Solutions To Questions, American Bond Options Page 1417 Chapter 29: Solutions To Questions, American Bond Options Page 1418 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures 2. Either the put or call will be exercised in 1992. Thus, the certain cash flows are: Solutions to Bonds with Embedded Options 1
1. Part a. False. The price of a noncallable bond always exceeds the price of a comparable callable bond assuming the value of the implicit call option is positive. Given this price difference, the yield on a callable must always be greater than the yield on a noncallable bond. '89 '90 10 '91 10 '92 100 ........ . + 10
. .... ... ... ... ... ... ... ... ... ... '93 exercised '94 exercised call or put price B = 10 e.08 + 10 e.092 + 110 e.103 99.07 = Part b. True. The value of a zero will always be less than its par value as long as interest rates are positive. The call option should not be exercised if the call price exceeds the value of the underlying bond. Part c. True. Call options on noncoupon bearing bonds are worth more alive than dead as long as the exercise price is constant. Part d. False. The desirability of exercising the call option will depend on the schedule of call prices. With changing exercise prices, early exercise may be desirable. Part e. False. We know the value of the callable bond must be less than 850. 850 is the upper bound on the value of callable bond assuming value of implicit call option is zero. You should not pay 900 for a security worth less than 850.00. Part f. False. Call the bond. Since the option is European and ready to expire, you must exercise now or never. Since the option is "inthemoney," you should call the bond. Part g. True. An increase in the default risk should lead to an increase in the default premium. This will lower the value of the bond. Holding the call prices fixed, a lower value for the bond will tend to discourage calling the bond.
Chapter 29: Solutions To Questions, Bonds with Embedded Options 1 Page 1420 Chapter 29: Solutions To Questions, Bonds with Embedded Options 1 Page 1419 Adventures in Debentures Copyright c 2004 by Michael R. Gibbons Adventures in Debentures Solving 2 equations in 2 unknowns Solutions to Bonds with Embedded Options 2
1. "ExCoupon" Prices for Callable Bond: .. ... ... ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... .. .. ... ... .. .. .. .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... N1 = .3445 NNC = .7081 Cost of replicating callable bond = N1 B1 + NNC BNC = 1,093.74 1,000 Since value of callable bond if not called (1,093.74) exceeds its value if called (1,070), borrower will call. Thus, value of bond in year 0 is $1,070.00. 1,070 (called!) .. .. ... ... .. ... ... ... .. .. ... ... ... ... ... ... . . ... ... ... ... ... ... . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... 1,030.17 1,000 1,060 (called!) 2. Part a. In year 3, under all scenarios the third bond pays $109 to the person who bought it in year 2. Thus, to value the third bond as of year 2, we need to discount the riskless payment of $109 at the appropriate discount rate and then compare the price to the call and put prices. That is, In up scenario of year 2, B = .8839 109 = 96.3451. Since 96.3451 < 99.0000, the bondholders will put the bond back to the corporation. Thus, the bond is worth 99.0000. In the middle scenario of year 2, B = .9324 109 = 101.6316. Since 99 < 101.6316 < 103, the bond will not be put or called. In the down scenario of year 2, B = .9836 109 = 107.2124. Since 107.2124 > 103, the corporation will call the bond at 103. Thus the bond is worth 103. Now we need to compute the value of the third bond in the down scenario of year 1. 1,000 Value in year 1 in the upstate is the same as the value for noncallable bond. At this point the callable bond is the same as noncallable because bond will not be called. Value in year 1 in the downstate if bond not called is the same as noncallable. This value is 1,072.30. However, borrower can minimize market value of debt by calling bond at 1,060.00. Thus, bond is called, and its value is 1,060.00. Value in year 0. First calculate value if bond not called: Use a portfolio of 1 period bonds and the noncallable coupon bond to replicate payoff of callable bond up: 1,000 N1 + (1,030.17 + 150)NNC = (1,030.17 + 150) down: 1,000 N1 + (1,072.30 + 150)NNC = (1,060 + 150) Given the continuously compounded yield is 9.0%, the price of a 1 year zero with face value = 1,000.00 is $913.93. Thus, B1 = 913.93 BNC = 1,099.99
Page 1421 Year 1 down
.. .. ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. .. ... ... .. .. ... ... . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... Year 2 101.6316 ? 103.0000
Page 1422 Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 Adventures in Debentures Adventures in Debentures To find ?, 1N1 + (101.63 + 9.00)NB = (101.6316 + 9) 1N1 + (107.21 + 9)NB = (103 + 9) NB = 0.2452 N1 = 83.5015 Cost as of Year 1 of the replicating portfolio = .9576 N1 + 107.35 NB = 106.28677 Part b. Value at t = 0: 1N1 + (971.23 + 131.72)Nnc = (971.23 + 131.72) N1 = 57.7029 1N1 + (1098.34 + 131.72)Nnc = (1091.69 + 131.72) Nnc = 0.9477 determined in part a. above Cost of Synthetic = .89360 N1 + 1060.49 Nnc = 1056.57 . So corporation should call bond at 106. 106 is the value at ? Part b. Summary of the optimal exercise of put and call:
.. .. .. .. ... ... ... ... ... ... .. ... ... ... . .. ... ... .. .. ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... . .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... exercise put feature NOTE: The strategy derived in part a and the valuation in part b are similar to what we will encounter with the timing option for Treasury bond futures. Part c. We observe (price of non callable) > (price of callable with notice) > (price of callable without notice). This is expected, for notification makes the call option less desirable to borrower and more desirable to lender (who will then pay more for the issue at t = 0). Notification may force the borrower to call the bond at t = 2 when rates are high. Part d. Value of costs associated with giving notice at t = 1: do nothing . ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... do nothing do nothing exercise call feature exercise call feature e.075 [1060 + 131.72] = 1105.61 3. Part a. Look at bottom scenario at t = 1:
Valuation of Not Giving Notice .. ... .. ... ... ... .. .. ... ... ... ... ... ... . ... ... ... .. ... ... ... ... ... ... .. .. ... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . Valuation of Giving Notice .. ... .. ... ... ... .. .. ... ... ... ... ... ... . ... ... ... .. ... ... ... ... ... ... .. .. ... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . .. .. ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... . ... ... . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . 1060 + 131.72 1060 + 131.72 t=2 1011.30 + 131.72 1045 + 131.72 t=1 1098.34 e.075 [1045 + 131.72] = 1091.69 1090.07 + 131.72 1045 + 131.72 Clearly giving notice at t = 1 and calling the bond at t = 2 is not desirable when the call price = 1060. (Note a call would be done at t = 2 if notice were not required at t = 1.) We now compare 1105.61 with 1098.34 (see part a) to verify that it is cheaper not to call! Part e. The value of the callable bond which requires notice and has a call price of 1060 is the same as the value of the noncallable. The call option is never used and has no value. Hence, in Year 0 this type of callable bond should sell for 1060.49. We know the PV of the cost of not giving notice is 1098.34 because the cost is the same as a noncallable bond. If we do give notice at t = 1 and call the bond at t = 2, the PV of the cost of this strategy is 1091.69. Since the cost of the latter is less than the former, we ought to notify at t = 1 and call at t = 2.
Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 Page 1423 Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 Page 1424 Adventures in Debentures Adventures in Debentures 4. Part a. I would expect the price of the PIK bond to be less than the price of a comparable nonPIK bond. A PIK bond provides an option to the issuer to make future coupon payments in kind. These inkind payments will have less value than cash if the issuer makes the proper decision. (A PIK bond may have a value equal to a nonPIK alternative if the PIK bond always sells for par value or greater.) Part b. A PIK bond provides a putlike feature to the issuer of the bond. In contrast, a callable bond provides a call option (not a put option) to the issuer of the bond. Also a putable bond provides a put option to the holder of the bond, not the issuer. Thus, a PIK bond is not like a callable or a putable bond. Part c. To find B2 , Part e. The following tree diagram will help keep track of the necessary calculations. The numbers in the tree are the prices of the callable PIK bond. These prices are not affected by the call provision, and they are the same as the noncallable PIK from part c above. Year 0 Year 1 Year 2
. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . ... ... ... ... ... ... ... ... . . ... ... ... ... . . ... ... ... ... .. .. ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . .. ... ... ... ... ... ... ... ... ... ... . . ... ... ... ... . ... ... ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . Year 3 1,130 B2 = 1,130 .86071 = 972.60 . C To find B1 : 1 N1 + .86071 N2 = (.97260 130) + 972.60 1 N1 + .92774 N2 = 130 + 1,048.35 Cost of this synthetic To find B0 : 1 N1 + .82903 N2 = (.90722 130) + 907.22 1 N1 + .89360 N2 = 130 + 1,028.83 Cost of this synthetic N1 = 691.08 N2 = 2,070.18 N1 = 80.62 N2 = 1,183.23 .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . 907.19 B .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . ... ... ... ... ... ... ... ... .. . ... .. .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . . . ... ... ... ... ... ... ... ... ... . . ... ... ... ... . . ... ... ... ... .. ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . 902.33 1,130 972.60 1,130 A = B1 = .89360 N1 + .80862 N2 = 1,028.83 . 1,130 To compute the price of a callable PIK bond at A: MIN{1,048.35, 980} = 980 . = B0 = .86071 N1 + .75801 N2 = 974.40 . To compute the price of a callable PIK bond at B: 1 N1 + .86071 N2 = (.97260 130) + 972.60 1 N1 + .92774 N2 = 130 + 980 Cost of this synthetic
Page 1425 Part d. We need to solve for the coupon yield such that the PIK bond sells for par value when rates are the highest. Thus, look at the top scenario in year 2. Find a coupon yield, k, such that [(1 + k)1,000] .79852 = 1,000 k 25.23% . =
Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 N1 = 958.28 N2 = 163.54 = B1 = .89360 N1 + .80862 N2 = 988.56 . Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 Page 1426 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons To compute the price of a callable PIK bond at C: 1 N1 + .82903 N2 = (130 .90719) + 907.19 1 N1 + .89360 N2 = (130 .98856) + 988.56 Cost of this synthetic N1 = 154.95 N2 = 1,423.48 = B0 = .86071 N1 + .75801 N2 = 945.65 . Solutions to Floating Rate Notes
1. Part a. This is a perfect floater. Perfect floaters sell for face value on reset dates. Thus, the market value is $5000. Part b. On date 2, the floater will pay .0697 5000 and can be sold in secondary market for $5000 on year 2. Thus 5000(1.0697) = 5048.52 . (1.08).75 Part c. We can replicate this floater with a margin using an FRN without a margin and an annuity that pays .02 5000 = 100 for years 1, 2, 3, 4. The floater without a margin has a market value equal to face value (or $5000). The annuity plus $5000 is: 1 1 1 1 + + + = 5304.91 . 5000 + 100 1.1052 (1.1133)2 (1.1196)3 (1.1247)4 Part d. We can replicate this floater by using the same instruments as in part c above. Further we need to short a security that pays 5% 5000 in year 1 (note we need 5% not 3% because the annuity in part c pays an extra 2% in year 1 which we must eliminate). Thus, using the cost of the synthetic in part c and adjusting it for the 5% 5000 payment in year 1: (.02 + .03)5000 = 5078.70 . 5304.9  1.1052 Part e. To find the appropriate "teaser" rate (^), we need to solve: r (.02 + r) 5000 ^ = 5000 5304.90  1.1052 r = 4.7395% . ^ Chapter 29: Solutions To Questions, Bonds with Embedded Options 2 Page 1427 Chapter 29: Solutions To Questions, Floating Rate Notes Page 1428 Adventures in Debentures Adventures in Debentures Thus, we need to set the coupon yield for the first coupon payment to:
0 r01 3. Part a. To value the inverse floater, we need to find its perfect substitute. The following table illustrates how to construct a perfect substitute. (10 1,000) + (10 1,000 29 r29,30 ) 10,000 + 10,000(.2  29 r29,30 ) 2. Part a. Summary of value of FRN plus coupon paid to holder from prior period: 1,066.50 1.0653 + 92.20 10,000(.20  0 r0,1 ) 1,000 + 66.50 29 ? ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . ... ... ... ... ... ... . . ........ ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 1,000 + 118.40 1,000 + 66.50 .20 10,000 1,118.40 1.1238 + 92.20 . ... ... ... ... ... ... ... ... ... ... ... ... ... ... . ... ... ... ... ... ... ... ... . . ........ ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . .. ... ... ... ... ... ... ... ... ... ... ... ... .. .. ... ... ... ... ... ... ... ... . . ........ ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 1,000 + 118.40 10 1,000 28 r28,29 10,000(.2  28 r28,29 ) 10 1,000 0 r0,1 1N1 +
1 1.0653 N2 = 1,066.50 1.0653 + 92.20
1 1.0942 N2 = 121.3877
1 (1.0922)2 10 1,000 10 51.69 10 30 yr. zero 18,648.83 30 yr. constant coupon bond Inverse Floater Part b. Units FRN =  + 92.20  1,066.50 + 92.20 1.0653 [ln(1.1238)  ln(1.0653)] 100 = 1.1095475
1,118.40 1.1238 Chapter 29: Solutions To Questions, Floating Rate Notes Page 1429 Chapter 29: Solutions To Questions, Floating Rate Notes 10 1 1 30 yr. floater Security NOTE: This question and solution illustrate that the valuation of an imperfect floater may require a dynamic structured strategy. Such an approach is less general, for it requires the stronger assumptions implicit in the evolution of the interest rates as given by a tree diagram. In contrast, the pricing of perfect floaters is based on dynamic unstructured strategies, and the valuation does not require the stronger assumptions implicit in tree diagrams. Net 9,165.73 Cost as of Year 0 of the replicating portfolio = N1 + N2 = 996.82283 ? 0 1 0 1N1 + 1 1.1238 N2 = 1,118.40 1.1238 + 92.20 N1 = 979.3795 .20 10,000 To find "?" in the above tree, we need to solve: 10,000(.2  0 r0,1 ) 10,000(.2  28 r28,29 ) 0 10,000 + 10,000(.2  29 r29,30 )  4.7395% . 10,000 + (.20 10,000) 10 1,000 30 Page 1430 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons The net payoffs from the last 3 positions are summed and reported at the bottom of the prior table. The payoffs from the inverse floater are given at the top of the prior table. Since the net payoffs at the bottom match those at the top of the table, we know that 3 positions when combined in a portfolio provide a perfect substitute for the inverse floater. Since this perfect substitute cost 9,165.73, the inverse floater should cost the same amount. Part b. Security 30 yr. constant coupon Price 18,648.83 $ 1,549.58 1 P V [1,000 (1 + 0 r0,1 )] 100 = 10 30 P V (1,000) 100 30 51.69 = 100 = 15.507 $ 8.31% Solutions to Interest Rate Swaps
1. Part a. The appropriate fixed rate for this simple swap is just the par yield (annually compounded) on a 5 year constant coupon bond with annual coupon payments. This par yield (k5 ) is: k5 = 1  d5 d1 + d2 + d3 + d4 + d5 1  .6070 = .9231 + .8424 + .7613 + .6822 + .6070 = 10.2987% 10.30% 30 yr. floater 1,000.00 1% 30 yr. zero 51.69 30% Part b. Value of swap in up scenario in Year 1 from = 1,000  [(102.9715 .9048) + (102.9715 .8114) A's perspective + (102.9715 .7222) + (1,102.9715 .6387)] = 44.446397 Value of swap in down scenario in Year 1 from = 1,000  [(102.9715 .9418) + (102.9715 .8746) A's perspective + (102.9715 .8025) + (1,102.9715 .7287)] =  73.407393 Swaption: .. ... .. .. ... ... ... ... ... ... .. ... ... ... . .. ... ... ... ... ... ... .. .. . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... Inverse Floater = 30 yr. Const. Coup.  (10 Floater ) + 10 30 yr. Zero = 1,549.58  (10 10) + (10 15.507) = 1,604.65 Inverse
Floater = 1,604.65 Inverse Floater = = 17.51%. 9,165.73 9,165.73 44.446397 Notice how the implicit leverage of the 30 year inverse floater has increased its risk relative to the 30 year constant coupon bond and the floating rate note. ? 0 Year 1
Page 1432 Year 0
Chapter 29: Solutions To Questions, Floating Rate Notes Page 1431 Chapter 29: Solutions To Questions, Interest Rate Swaps Adventures in Debentures Adventures in Debentures When the value of the swaption is 0, the owner will not exercise the option. To find ?: 1N1 + .9048 N2 = 44.446397 1N1 + .9418 N2 = 0 N1 = 1,131.3410 N2 = 1,201.2540 (cost of replicating portfolio) = .9231 N1 + .8424 N2 = 32.40452 value of $100,000. (Note the face value for both instruments nets out to zero.) Thus, we need to find a fixed rate such that 128,030.41 = 100,000 k = 17.36% . k k k 1+k k + + + + 1.12 (1.11)2 (1.105)3 (1.10)4 (1.0975)5 Part c. In this case, Americanstyle swaption has same value as Europeanstyle swaption in part a. Exercising the swaption in year 0 implies the owner of swaption receives party A's position which has a present value of approximately zero, but early exercise in year 0 means you give up a swaption worth 32.40452. Thus, early exercise is not desirable. Since the early exercise feature of an American style swaption is not valuable, the value of an American style swaption is also 32.40452 in year 0. 2. See the table that follows to determine the dynamic (unstructured) synthetic. Part a. $ = 1 PV(120 + 2000) 2 PV(2000 + 3000) + 100 100 3 PV(3000 + 2000) 4 PV(1000  2000) + + 100 100 1 1 2120 2 1000 3 1000 4 1000 = + + + 100 (1.12) (1.11)2 (1.105)3 (1.10)4 = 1439.44 , where "PV" means present value. Part b. Clearly, the table gives the value of a synthetic roller coaster floater as $1,280.30. Part c. The synthetic version of this 5year swap can be created by issuing a roller coaster floater with a face value of $100,000 and buying a fixed rate bond with a face
Chapter 29: Solutions To Questions, Interest Rate Swaps Page 1433 Chapter 29: Solutions To Questions, Interest Rate Swaps Page 1434 Adventures in Debentures 1,000(1 + 4 r4,5 ) receive 1,000 receive and reinvest  1,000(1 + 4 r4,5 ) at 4 r4,5 receive 1,000(1 + 4 r4,5 ) Adventures in Debentures Copyright c 2004 by Michael R. Gibbons 5 receive 2,000(1 + 3 r3,4 ) 1,000 2 3 r3,4 receive 1,000 2 3 r3,4 Solutions to Floating Rate Notes with Embedded Options
1. To value this option on the yield, the following tree provides a useful summary: Period 0 Period 0.25
. . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . ... .. .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . .. .. ... ... ... ... ... ... ... ... . . ... ... ... ... . ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . pay 2,000 4 receive 3,000(1 + 2 r2,3 ) 1,000 3 2 r2,3 receive 1,000 3 2 r2,3 Period 0.50 0 Period 0.75 receive 2,000 and reinvest at 3 r3,4 pay 3,000 3 receive 2,000 receive and reinvest  2,000(1 + 1 r1,2 ) at 1 r1,2 1,000 2 1 r1,2 receive 1,000 2 1 r1,2 B receive 3,000 and reinvest at 2 r2,3 pay 2,000 2 .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . .. ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . 0 0 A 572,500.00 = . . . . 548,228.86 = ..................................................... 100,000,000 572,500 .957605 [.2000  .1771]1/4 receive 120 1,000 0 r0,1 receive 120 1 To find the cash payment in arrears, we know payment = Notional Amount Strike Rate y
1/ 4 3,000 (1.105)3 2,000 (1.11)2 2,000 (1.10)4 2,000 (1.105)3 3,000 (1.11)2 receive receive receive pay 1,280.3041 1,000 (1.1)4 0 pay pay 2,000 1.12 120 1.12 where y is the relevant annualized rate with quarterly compounding. That is, y is the solution: .957605 = 1 1 + y/4 y=4 1  1 17.71% . = .957605 N1 = 450,339.76 NB = 430,253.38 pay pay buy 1year zero with Face = 2,000 pay issue 3year zero with Face = 3,000 issue 2year zero with Face = 2,000 buy 4year zero with Face = 1,000 buy 2year zero with Face = 3,000 buy 3year zero with Face = 2,000 issue 4year zero with Face = 2,000 buy 1year zero with Face = 120 To find the value at point A on the above tree:
Net 100 N1 + (99.7485 + 4.92) NB = 0 100 N1 + (101.0227 + 4.92) NB = 548,228.62 Cost of this Synthetic = 95.4412 N1 + 100.7250 NB = 356,304.69 . Chapter 29: Solutions To Questions, Interest Rate Swaps Page 1435 Chapter 29: Solutions To Questions, Floating Rate Notes with Embedded Options Page 1436 Adventures in Debentures Adventures in Debentures To find the value at point B on the above tree: 100 N1 + (98.9177 + 4.92) NB = 0 100 N1 + (100.7250 + 4.92) NB = 356,304.09 Cost of this Synthetic N1 = 204,713.44 NB = 197,147.51 Thus, buy two units of fixed rate obligation and short one unit of floating rate note to replicate one unit of the inverse floater. Part c. Using the synthetic from part b above and recognizing the present value of the floating rate note is 100, we need to find k such that the present value of the fixed rate obligation is 200. Thus, 200 = 200(k/2)(.97) + 200(k/2)(.93) + 200(1 + k/2)(.89) k = 7.8853% . = 95.1229 N1 + 99.9919 NB = 240,218.02 . 2. Part a. To find the value of the inverse floater, we need to find the value of a synthetic inverse floater using the swap and a 3 year zero. The following table shows that a swap (receive fixed and pay floating) with a notional amount of $100 plus a 3 year zero with a face value of 100 is a synthetic inverse floater with face value of 100. Part d. If we combine an inverse floater with the appropriate cap, such a portfolio replicates the payoffs of a protected inverse floater. We need a cap with a strike rate of 7.5% for a notional amount of 100 (not 100,000). We know the market value of such a cap is 1,075/1,000 = $1.075. From part b, we can value the synthetic version of an unprotected inverse floater and then add the cost of the cap:
P V of two units of Fixed Rate = 200 = .075 2 (.97) + 200 .075 2 (.93) + 200 1 + .075 2 (.89) 0 (1) Inverse Floater (2) Swap: Receive Fixed = k; Pay Floating; Notional Amt = 100 (3) Buy a 3 year zero with Face = 100 1 100(k  0 r0,1 ) 100(k  0 r0,1 ) 2 100(k  1 r1,2 ) 100(k  1 r1,2 ) 3 [100 + 100(k  2 r2,3 )] 100(k  2 r 2,3 ) 198.925 0 0 100 Minus P V of one unit = 100.00 of a Floating Rate Note Plus Cost of Cap Total P V of Synthetic = +1.075 Thus, cash flows in row 1 equal cash flows in row 2 plus row 3. The value of the snythetic = value of swap + value of 3 year zero = 0 + .89 100 = 89 . Part b. The following table demonstrates the appropriate synthetic that replicates the inverse floaters: = $100.00 0 (1) Inverse Floater (2) Short one unit of floating rate note (3) Buy two units of fixed rate 1 100(k  0 r0,1 ) 100(0 r0,1 ) 200(k/2) 2 100(k  1 r1,2 ) 100(1 r1,2 ) 200(k/2) 3 [100 + 100(k  2 r2,3 )] 100(1 + 2 r2,3 ) 200(1 + k/2) 3. First, we need to analyze the cap. The cap is a portfolio of call options on yields, where each call option has a strike rate of 10% and a notional amount of 500,000 (not 100,000). Thus, the cap is worth in the up state of period 1: 5 (191.73 + 149.43) . In the down state of period 1, the cap is worthless.
Chapter 29: Solutions To Questions, Floating Rate Notes with Embedded Options Page 1438 Chapter 29: Solutions To Questions, Floating Rate Notes with Embedded Options Page 1437 Adventures in Debentures Adventures in Debentures Copyright c 2004 by Michael R. Gibbons The caption matures on period 1 with an exercise price of 705.80, its value:
. . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. 5(191.73 + 149.43)  705.80 = 1,000 ? Solutions to Home Mortgages
1. Part a. payment = y(1 + y)2 .121 100,000 = 100,000 = 57,619.05. (1 + y)2  1 .21 0 Period 1 Period 0 For this particular problem the payoffs in period 1 are very similar for this caption, the cap, and the two call options on yields, for all four are worthless in period 1 in the down state. This makes it easy to replicate the payoff of the caption using any one of the other three. For example, without using two equations in two unknowns, we can replicate the payoffs from the caption using the shortterm call option on the yield: 1,000 = 5.216 units of the call option will 191.73 have same value on period 1 as the caption. Hence, on period 0 the value of the caption is 5.216 105.71 = 551.38 . Period 1 2 Mortgage Balance at Start of Period 100,000 52,380.95 Interest $10,000 5,238.10 Payment 57,619.05 57,619.05 Mortgage Balance at End 52,380.95 0 If borrower prepays mortgage in period 1, he must pay lender 52,380.95 (plus accrued interest). Alternatively, one can solve the problem with the following equation: Mt = M0 (1 + y/m)mT  (1 + y/m)t . (1 + y/m)mT  1 Using the last equation for this specific case: M1 = 100,000 (1 + .10/1)2  (1 + .10/1)1 = 52,380.95 . (1 + .10/1)2  1
Page 1440 Chapter 29: Solutions To Questions, Floating Rate Notes with Embedded Options Page 1439 Chapter 29: Solutions To Questions, Home Mortgages Adventures in Debentures Adventures in Debentures Part b. Consider how to value this mortgage: Year 0 Year 1
. .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. .. . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . .. .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... . ... ... ... .. .. . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . 2. Part a. Value of IO: Year 2 57,619.05
. ... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... . ... ... ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . . .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . . ... ... ... ... ... ... ... ... . . ... ... ... ... . . ... ... ... ... . . ... ... ... ... ... ... ... ... . .......... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . 5,238.10 10,000 + (.86742 5,238.10) 98,195.07 . ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... . .. ... ... . . ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . .86742 57,619.05 = 49,979.92 < 52,380.95 Don't Prepay! 5,238.10 10,000 57,619.05 prepaid in year 1 .94387 57,619.05 = 54,384.89 > 52,380.95 Prepay! Value the IO in year 0 by finding the value of a synthetic that generates the same value (including cash flows) in year 1: up: 1 N1 + .86742 N2 = 10,000 + (.86742 5,238.10) = 14,543.63 57,619.05 Cost: .90483 N1 + .81147 N2 = 11,578.46 down: 1 N1 + .94387 N2 = 10,000 N1 = 66,096.74 N2 = 59,432.70 1N1 + .86742 N2 = 49,979.92 + 57,619.05 N1 = 80,356.31 N2 = 31,406.54 1N1 + .94387 N2 = 52,380.95 + 57,619.05 .90484 N1 + .81147 N2 = 98,195.07 Points = 100,000  98,195.07 = 1804.93 . Part b. Clearly, (Value of Mortgage) = (Value of IO) + (Value of PO) Part c. From part b we know the value of the callable annuity = 98,528.38. The value of noncall annuity (.90483 57,619.05) + (.81147 57,619.05) = 98,891.58. Thus, the value of the implicit option = 98,891.58  98,195.07 = 696.51. so (Value of PO) = 98,195.07  11,578.46 = 86,616.61 . Part c. IO =  Note t rt+1 =  ln t dt+1 . Chapter 29: Solutions To Questions, Home Mortgages Page 1441 14,543.63  10,000 14.22  5.78 = 538.34 Chapter 29: Solutions To Questions, Home Mortgages Page 1442 Adventures in Debentures Adventures in Debentures Part d. PO =  20,448.54 0.00 0.00 0.00 0.00 0.00 0.00 0.00 3 Total Cash Inflow 28,940.53 22,442.99 Amortization Schedule for Tranche A. (Assumes Original Mortgage Not Prepaid.) Princ Pd By Int from Z 2,926.05 0.00 Princ Pd By Orig Mort 21,625.41 20,448.54 4,389.07 1,994.45 Beginning Balance Interest 45,000.00 20,448.54 Yr 1 2 Chapter 29: Solutions To Questions, Home Mortgages Page 1443 Chapter 29: Solutions To Questions, Home Mortgages 4 0.00 0.00 0.00 0.00 0.00 3. Part a. The sixth column in the following three tables provides the total cash flow received by each tranche. Ending Balance 0.00 (49,979.92  14,543.63)  (52,380.95  10,000) 14.22  5.78 = 822.83 Page 1444 Adventures in Debentures Amortization Schedule for Tranche B. (Assumes Original Mortgage Not Prepaid.) Yr 1 2 3 4 0.00 0.00 0.00 0.00 18,502.46 1,804.64 18,502.46 0.00 20,307.10 0.00 25,000.00 2,438.37 3,286.10 3,211.44 8,935.91 25,000.00 2,438.37 0.00 0.00 2,438.37 25,000.00 18,502.46 0.00 0.00 Chapter 29: Solutions To Questions, Home Mortgages Beginning Balance Interest Princ Pd By Orig Mort Princ Pd By Int from Z Total Cash Inflow Ending Balance Page 1445 Adventures in Debentures Amortization Schedule for Tranche Z. (Assumes Original Mortgage Not Prepaid.) Yr 1 2 3 4 Beginning Balance 30,000.00 32,926.05 36,137.49 28,590.35 Interest 2,926.05 3,211.44 3,524.67 2,788.56 Interest Paid 0.00 0.00 3,524.67 2,788.56 Princ Pd By Orig Mort 0.00 0.00 7,547.14 28,590.34 Total Cash Inflow 0.00 0.00 11,071.81 31,378.90 Ending Balance 32,926.05 36,137.49 28,590.35 0.00 Chapter 29: Solutions To Questions, Home Mortgages Page 1446 Adventures in Debentures Adventures in Debentures Part b. Before calculating the market values for tranche A, lets write down the cash flows generated by the security. The relevant cash flows are the first number at each node: 20,448.54(1 + .0975349) = 22,442.99
. .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... .. ... ... ... .. .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. The second number (when it appears) at each node in the previous tree is the market value of tranche A. In the up scenario in year 1, the market value of tranche A is straightforward since the cash flows in year 2 are the same for both the up and down scenario. We only have to discount those cash flows by the riskless 1year rate as of the up scenario at year 1. (This rate is 10.83%.) To find the market value of tranche A as of year 0, we need to find a perfect substitute. For my perfect substitute, I held 1year paper and the original mortgage, which is the collateral of this CMO. Hence: Entire principal of Tranche A is eliminated! 28,940.53
. .. ... ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 22,442.99 1.1083 = 20,249.92 0 45,468.35 20,448.54(1 + .0975349) = 22,442.99 Entire CMO is eliminated! up: 1 N1 + (76,252.06 + 31,378.90)NM = (20,249.92 + 28,940.53) down: 1 N1 + (78,374.59 + 31,378.90)NM = 49,389.07 1 N1 = 39,118.6671 N1 + 100,000 NM = 45,468.35 cost: 1.0833 NM = 0.0936 45,000(1 + .0975349) = 49,389.07 Entire CMO is eliminated! Year 0 Year 1 Year 2 If you want additional practice for problems like this one, you should confirm the valuation of tranche Z reported in the tree diagram within the question. Part c. Since we know the market values of the original mortgage, tranche A, and Tranche Z, the market value of tranche B can be computed by simple subtraction. That is, The cash flow in the up scenario in year 1 was given to you in the question; 28,940.53 = 4,389.07 + 21,625.41 + 2,926.05. The cash flow in the down scenario in year 1 reflects that the entire CMO is prepaid. Thus, the cash flow to tranche A is the outstanding principal (45,000) plus interest on that principal (45,000 .0975349). For both scenarios in year 2, tranche A is clearly paid off. From the table in the question, we see that the principal portion of the second payment, i.e., 31,378.90  7,644.26 = 23,734.64, is greater than the outstanding principal on tranche A (20,448.54). Thus, holders of tranche A on year 2 receive principal (20,448.54) plus interest on that principal (20,448.34 .0975349).
Chapter 29: Solutions To Questions, Home Mortgages Page 1447 Mkt. Value of Tranche B = Mkt. Value of  Mkt. Value of  Mkt. Value of Original Mortgage Tranche A Tranche Z The following tree summarizes these calculations for the market value of tranche B.
Chapter 29: Solutions To Questions, Home Mortgages Page 1448 Adventures in Debentures Adventures in Debentures 0 (entire CMO prepaid) Vasicek Delta 0 (entire CMO prepaid) Part d. The following table summarizes the relevant delta calculations that are needed to compute the omega values that follow. Mort. = B = 465.4671 = 0.47% 100,000 A = Z = 43.5570 = 0.10% 45,468.35 122.2171 = .49% 24,983.48 299.6923 = 1.01% 29,548.17 Price as of Year 0 100,000.00 45,468.35 24,983.48 Original mortgage Security Tranche A Tranche B Chapter 29: Solutions To Questions, Home Mortgages Page 1449 Chapter 29: Solutions To Questions, Home Mortgages Tranche Z 29,548.17 You can verify that Mort. A + B + Z . Thus, the creation of the CMO did = not eliminate any risk; the CMO merely changes the allocation of the risk. For those investors who wish to participate in the mortgage market but who wish to have little risk, such an investor should purchase tranche A. For those investors who are willing to take more risk, such investors should purchase tranche Z. (20,249.92 + 28,940.53)  (0 + 49,389.07) = 43.5570 10.28  5.72 100,000.00 45,468.35 29,548.17 = 24,983.48    (31,559.45 + 0)  (0 + (30,000 1.0975349)) = 299.6923 10.28  5.72 . . .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... . ... ... .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . Note: the denominator in each of the above calculations converts from annually compounded to continuously compounded. (76,252.06 + 31,378.90)  (78,374.59 + 31,378.90)  = 465.4671 10.28  5.72 76,252.06 20,249.92 31,559.45 = 24,442.69 . ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... .. ... ... ... . . .. .. ... .. ....... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . 52,079.65  0  34,168.67 = 17,910.98 (24,442.69 + (.0975349 25,000))  (0 + (25,000 1.0975349)) = 122.2171 10.28  5.72 Remember the market value reflects the value of the remaining cash flows Page 1450 ...
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This note was uploaded on 03/27/2012 for the course FNCE 235 taught by Professor Roussanov during the Spring '09 term at UPenn.
 Spring '09
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